programming c++

Cách đảo ngược một số trong c ++ mà không cần sử dụng vòng lặp

Reverse of no. is 26542Reverse of no. is 12345607Input: num (1) Initialize rev_num = 0 (2) Loop while num > 0 (a) Multiply rev_num by 10 and add remainder of num divide by 10 to rev_num rev_num = rev_num*10 + num%10; (b) Divide num by 10 (3) Return rev_num85Input: num (1) Initialize rev_num = 0 (2) Loop while num > 0 (a) Multiply rev_num by 10 and add remainder of num divide by 10 to rev_num rev_num = rev_num*10 + num%10; (b) Divide num by 10 (3) Return rev_num64 Reverse of no. is 12345639Input: num (1) Initialize rev_num = 0 (2) Loop while num > 0 (a) Multiply rev_num by 10 and add remainder of num divide by 10 to rev_num rev_num = rev_num*10 + num%10; (b) Divide num by 10 (3) Return rev_num77 Input: num (1) Initialize rev_num = 0 (2) Loop while num > 0 (a) Multiply rev_num by 10 and add remainder of num divide by 10 to rev_num rev_num = rev_num*10 + num%10; (b) Divide num by 10 (3) Return rev_num24Reverse of no. is 265458Input: num (1) Initialize rev_num = 0 (2) Loop while num > 0 (a) Multiply rev_num by 10 and add remainder of num divide by 10 to rev_num rev_num = rev_num*10 + num%10; (b) Divide num by 10 (3) Return rev_num83 Reverse of no. is 12345610

Reverse of no. is 26542______110Input: num (1) Initialize rev_num = 0 (2) Loop while num > 0 (a) Multiply rev_num by 10 and add remainder of num divide by 10 to rev_num rev_num = rev_num*10 + num%10; (b) Divide num by 10 (3) Return rev_num83Input: num (1) Initialize rev_num = 0 (2) Loop while num > 0 (a) Multiply rev_num by 10 and add remainder of num divide by 10 to rev_num rev_num = rev_num*10 + num%10; (b) Divide num by 10 (3) Return rev_num64 Input: num (1) Initialize rev_num = 0 (2) Loop while num > 0 (a) Multiply rev_num by 10 and add remainder of num divide by 10 to rev_num rev_num = rev_num*10 + num%10; (b) Divide num by 10 (3) Return rev_num24

Reverse of no. is 12345653

Reverse of no. is 12345654Input: num (1) Initialize rev_num = 0 (2) Loop while num > 0 (a) Multiply rev_num by 10 and add remainder of num divide by 10 to rev_num rev_num = rev_num*10 + num%10; (b) Divide num by 10 (3) Return rev_num64 Input: num (1) Initialize rev_num = 0 (2) Loop while num > 0 (a) Multiply rev_num by 10 and add remainder of num divide by 10 to rev_num rev_num = rev_num*10 + num%10; (b) Divide num by 10 (3) Return rev_num51

Reverse of no. is 12345661Reverse of no. is 12345662

Reverse of no. is 12345663

C#

Reverse of no. is 12345664

Reverse of no. is 265405

Reverse of no. is 1234562 Input: num (1) Initialize rev_num = 0 (2) Loop while num > 0 (a) Multiply rev_num by 10 and add remainder of num divide by 10 to rev_num rev_num = rev_num*10 + num%10; (b) Divide num by 10 (3) Return rev_num99

Reverse of no. is 1234563598 Reverse of no. is 1234563599

Reverse of no. is 26542____052 Reverse of no. is 26541

Reverse of no. is 26542____07

Reverse of no. is 26542Reverse of no. is 26549 Reverse of no. is 26540

Reverse of no. is 26542____21 Reverse of no. is 26548

Reverse of no. is 26542Reverse of no. is 265416

Reverse of no. is 26547

Reverse of no. is 1234563595

Reverse of no. is 265421

PHP

Reverse of no. is 265452

Reverse of no. is 265423

Reverse of no. is 265461 Reverse of no. is 265462

Reverse of no. is 265426 Reverse of no. is 265427

Reverse of no. is 265428

Reverse of no. is 265429

Reverse of no. is 265455 Reverse of no. is 265439____057____058

Input: num (1) Initialize rev_num = 0 (2) Loop while num > 0 (a) Multiply rev_num by 10 and add remainder of num divide by 10 to rev_num rev_num = rev_num*10 + num%10; (b) Divide num by 10 (3) Return rev_num6Reverse of no. is 12345618 Reverse of no. is 265461Input: num (1) Initialize rev_num = 0 (2) Loop while num > 0 (a) Multiply rev_num by 10 and add remainder of num divide by 10 to rev_num rev_num = rev_num*10 + num%10; (b) Divide num by 10 (3) Return rev_num16

Input: num (1) Initialize rev_num = 0 (2) Loop while num > 0 (a) Multiply rev_num by 10 and add remainder of num divide by 10 to rev_num rev_num = rev_num*10 + num%10; (b) Divide num by 10 (3) Return rev_num6Reverse of no. is 12345618 Reverse of no. is 265426Input: num (1) Initialize rev_num = 0 (2) Loop while num > 0 (a) Multiply rev_num by 10 and add remainder of num divide by 10 to rev_num rev_num = rev_num*10 + num%10; (b) Divide num by 10 (3) Return rev_num16

Reverse of no. is 26542____051____057 Reverse of no. is 265453

Reverse of no. is 26542____061 Reverse of no. is 265456____057 Reverse of no. is 265458

Reverse of no. is 265459Reverse of no. is 265426Input: num (1) Initialize rev_num = 0 (2) Loop while num > 0 (a) Multiply rev_num by 10 and add remainder of num divide by 10 to rev_num rev_num = rev_num*10 + num%10; (b) Divide num by 10 (3) Return rev_num16

Reverse of no. is 26542Reverse of no. is 265426 Reverse of no. is 265464

Input: num (1) Initialize rev_num = 0 (2) Loop while num > 0 (a) Multiply rev_num by 10 and add remainder of num divide by 10 to rev_num rev_num = rev_num*10 + num%10; (b) Divide num by 10 (3) Return rev_num6Reverse of no. is 26549 Reverse of no. is 265461Input: num (1) Initialize rev_num = 0 (2) Loop while num > 0 (a) Multiply rev_num by 10 and add remainder of num divide by 10 to rev_num rev_num = rev_num*10 + num%10; (b) Divide num by 10 (3) Return rev_num16

Reverse of no. is 26547

Reverse of no. is 265490

Reverse of no. is 265457 Reverse of no. is 265492

________ 093 ________ 246 ________ 095

Reverse of no. is 265496Reverse of no. is 265439____057Reverse of no. is 265499

Reverse of no. is 265482

Reverse of no. is 265401

Javascript

Reverse of no. is 265402

Reverse of no. is 265485

Reverse of no. is 265437

Reverse of no. is 265487 Input: num (1) Initialize rev_num = 0 (2) Loop while num > 0 (a) Multiply rev_num by 10 and add remainder of num divide by 10 to rev_num rev_num = rev_num*10 + num%10; (b) Divide num by 10 (3) Return rev_num8

Reverse of no. is 265487 Reverse of no. is 265450

Reverse of no. is 265455 Reverse of no. is 265492

_______02____16200

Reverse of no. is 26542____16202

Reverse of no. is 26542____059

Reverse of no. is 26547

Reverse of no. is 265490

Reverse of no. is 26542____16218

Reverse of no. is 1234563595

Reverse of no. is 1234566220

Reverse of no. is 265434

Đầu raReverse of no. is 2654

Độ phức tạp về thời gian. O(log(n))
Không gian phụ. O(log(n)), trong đó n là số đầu vào.

Sử dụng đệ quy, không có biến phụ

C++

Reverse of no. is 1234566222

Reverse of no. is 1234566223

Reverse of no. is 1234562 Reverse of no. is 1234563 Reverse of no. is 1234564

Reverse of no. is 1234566227

Reverse of no. is 26547

Reverse of no. is 1234566240

Reverse of no. is 26542Reverse of no. is 26549 Reverse of no. is 1234566251

Reverse of no. is 26547

Java

Reverse of no. is 1234566279 Reverse of no. is 1234566280

Reverse of no. is 1234566279 Reverse of no. is 1234566282

Reverse of no. is 1234563598 Reverse of no. is 1234563599

_______02____16234

Reverse of no. is 26542Reverse of no. is 26549 Input: num (1) Initialize rev_num = 0 (2) Loop while num > 0 (a) Multiply rev_num by 10 and add remainder of num divide by 10 to rev_num rev_num = rev_num*10 + num%10; (b) Divide num by 10 (3) Return rev_num31Input: num (1) Initialize rev_num = 0 (2) Loop while num > 0 (a) Multiply rev_num by 10 and add remainder of num divide by 10 to rev_num rev_num = rev_num*10 + num%10; (b) Divide num by 10 (3) Return rev_num1____2400Input: num (1) Initialize rev_num = 0 (2) Loop while num > 0 (a) Multiply rev_num by 10 and add remainder of num divide by 10 to rev_num rev_num = rev_num*10 + num%10; (b) Divide num by 10 (3) Return rev_num351Input: num (1) Initialize rev_num = 0 (2) Loop while num > 0 (a) Multiply rev_num by 10 and add remainder of num divide by 10 to rev_num rev_num = rev_num*10 + num%10; (b) Divide num by 10 (3) Return rev_num16

Reverse of no. is 26542Reverse of no. is 265452 Input: num (1) Initialize rev_num = 0 (2) Loop while num > 0 (a) Multiply rev_num by 10 and add remainder of num divide by 10 to rev_num rev_num = rev_num*10 + num%10; (b) Divide num by 10 (3) Return rev_num418Input: num (1) Initialize rev_num = 0 (2) Loop while num > 0 (a) Multiply rev_num by 10 and add remainder of num divide by 10 to rev_num rev_num = rev_num*10 + num%10; (b) Divide num by 10 (3) Return rev_num24 Input: num (1) Initialize rev_num = 0 (2) Loop while num > 0 (a) Multiply rev_num by 10 and add remainder of num divide by 10 to rev_num rev_num = rev_num*10 + num%10; (b) Divide num by 10 (3) Return rev_num420

Reverse of no. is 26542____07

Reverse of no. is 26542Input: num (1) Initialize rev_num = 0 (2) Loop while num > 0 (a) Multiply rev_num by 10 and add remainder of num divide by 10 to rev_num rev_num = rev_num*10 + num%10; (b) Divide num by 10 (3) Return rev_num1 Input: num (1) Initialize rev_num = 0 (2) Loop while num > 0 (a) Multiply rev_num by 10 and add remainder of num divide by 10 to rev_num rev_num = rev_num*10 + num%10; (b) Divide num by 10 (3) Return rev_num428Input: num (1) Initialize rev_num = 0 (2) Loop while num > 0 (a) Multiply rev_num by 10 and add remainder of num divide by 10 to rev_num rev_num = rev_num*10 + num%10; (b) Divide num by 10 (3) Return rev_num24Input: num (1) Initialize rev_num = 0 (2) Loop while num > 0 (a) Multiply rev_num by 10 and add remainder of num divide by 10 to rev_num rev_num = rev_num*10 + num%10; (b) Divide num by 10 (3) Return rev_num16

Reverse of no. is 26542______21 Input: num (1) Initialize rev_num = 0 (2) Loop while num > 0 (a) Multiply rev_num by 10 and add remainder of num divide by 10 to rev_num rev_num = rev_num*10 + num%10; (b) Divide num by 10 (3) Return rev_num433Input: num (1) Initialize rev_num = 0 (2) Loop while num > 0 (a) Multiply rev_num by 10 and add remainder of num divide by 10 to rev_num rev_num = rev_num*10 + num%10; (b) Divide num by 10 (3) Return rev_num24Input: num (1) Initialize rev_num = 0 (2) Loop while num > 0 (a) Multiply rev_num by 10 and add remainder of num divide by 10 to rev_num rev_num = rev_num*10 + num%10; (b) Divide num by 10 (3) Return rev_num16

Reverse of no. is 26542______09 Input: num (1) Initialize rev_num = 0 (2) Loop while num > 0 (a) Multiply rev_num by 10 and add remainder of num divide by 10 to rev_num rev_num = rev_num*10 + num%10; (b) Divide num by 10 (3) Return rev_num441Input: num (1) Initialize rev_num = 0 (2) Loop while num > 0 (a) Multiply rev_num by 10 and add remainder of num divide by 10 to rev_num rev_num = rev_num*10 + num%10; (b) Divide num by 10 (3) Return rev_num1____2443Input: num (1) Initialize rev_num = 0 (2) Loop while num > 0 (a) Multiply rev_num by 10 and add remainder of num divide by 10 to rev_num rev_num = rev_num*10 + num%10; (b) Divide num by 10 (3) Return rev_num24Input: num (1) Initialize rev_num = 0 (2) Loop while num > 0 (a) Multiply rev_num by 10 and add remainder of num divide by 10 to rev_num rev_num = rev_num*10 + num%10; (b) Divide num by 10 (3) Return rev_num445

Reverse of no. is 26542____2463

Reverse of no. is 26547

Python3

Reverse of no. is 12345615 Input: num (1) Initialize rev_num = 0 (2) Loop while num > 0 (a) Multiply rev_num by 10 and add remainder of num divide by 10 to rev_num rev_num = rev_num*10 + num%10; (b) Divide num by 10 (3) Return rev_num470

Reverse of no. is 12345615 Reverse of no. is 12345635922

Reverse of no. is 12345635930

Reverse of no. is 265452 Reverse of no. is 12345635932____264Input: num (1) Initialize rev_num = 0 (2) Loop while num > 0 (a) Multiply rev_num by 10 and add remainder of num divide by 10 to rev_num rev_num = rev_num*10 + num%10; (b) Divide num by 10 (3) Return rev_num64 Reverse of no. is 12345635935Reverse of no. is 12345635936

C#

_______02____16234

Reverse of no. is 26542____09 Input: num (1) Initialize rev_num = 0 (2) Loop while num > 0 (a) Multiply rev_num by 10 and add remainder of num divide by 10 to rev_num rev_num = rev_num*10 + num%10; (b) Divide num by 10 (3) Return rev_num31____21Reverse of no. is 12345635962

Reverse of no. is 26542Reverse of no. is 265452 Reverse of no. is 12345635977

Reverse of no. is 26542____2012

Reverse of no. is 26547

Javascript

Reverse of no. is 265402

Reverse of no. is 265485

Reverse of no. is 265406

Reverse of no. is 265455 Input: num (1) Initialize rev_num = 0 (2) Loop while num > 0 (a) Multiply rev_num by 10 and add remainder of num divide by 10 to rev_num rev_num = rev_num*10 + num%10; (b) Divide num by 10 (3) Return rev_num021

Reverse of no. is 26547

Reverse of no. is 1234566240

Reverse of no. is 26542Reverse of no. is 26549 Reverse of no. is 12345635980

Reverse of no. is 26547

Reverse of no. is 265434

PHP

Reverse of no. is 265452

Reverse of no. is 265423

Reverse of no. is 265406

Reverse of no. is 265455 Reverse of no. is 12345635951____057Input: num (1) Initialize rev_num = 0 (2) Loop while num > 0 (a) Multiply rev_num by 10 and add remainder of num divide by 10 to rev_num rev_num = rev_num*10 + num%10; (b) Divide num by 10 (3) Return rev_num059

Reverse of no. is 26547

Reverse of no. is 1234566240

Reverse of no. is 265455 Reverse of no. is 1234566242____057Input: num (1) Initialize rev_num = 0 (2) Loop while num > 0 (a) Multiply rev_num by 10 and add remainder of num divide by 10 to rev_num rev_num = rev_num*10 + num%10; (b) Divide num by 10 (3) Return rev_num059

Reverse of no. is 26542Reverse of no. is 26549 Reverse of no. is 265457____216

Reverse of no. is 26547

Reverse of no. is 265490

Reverse of no. is 265457 Input: num (1) Initialize rev_num = 0 (2) Loop while num > 0 (a) Multiply rev_num by 10 and add remainder of num divide by 10 to rev_num rev_num = rev_num*10 + num%10; (b) Divide num by 10 (3) Return rev_num110

Reverse of no. is 265493 Reverse of no. is 1234566242____057Reverse of no. is 265499

Reverse of no. is 265401

Đầu raInput: num (1) Initialize rev_num = 0 (2) Loop while num > 0 (a) Multiply rev_num by 10 and add remainder of num divide by 10 to rev_num rev_num = rev_num*10 + num%10; (b) Divide num by 10 (3) Return rev_num3

Độ phức tạp về thời gian. O(log10n) trong đó n là số đầu vào đã cho.
Không gian phụ. O(log10n) cho không gian ngăn xếp đệ quy.

Sử dụng String trong java

Chúng tôi sẽ chuyển đổi số thành một chuỗi bằng StringBuffer sau này, chúng tôi sẽ đảo ngược chuỗi đó bằng phương thức reverse()

trường hợp góc

Đầu vào. 32100
Vì vậy, đối với đầu vào ở trên, nếu chúng ta cố gắng giải quyết vấn đề này bằng cách đảo ngược chuỗi, thì đầu ra sẽ là 00123
Vì vậy, để giải quyết tình huống này, một lần nữa chúng ta cần chuyển đổi chuỗi thành số nguyên để đầu ra của chúng ta là 123

C

Reverse of no. is 123456359

_______02____2137

Reverse of no. is 26542____2139

Reverse of no. is 26542____2141

Reverse of no. is 26547

Reverse of no. is 26542____2167

Reverse of no. is 26542Reverse of no. is 265452 Input: num (1) Initialize rev_num = 0 (2) Loop while num > 0 (a) Multiply rev_num by 10 and add remainder of num divide by 10 to rev_num rev_num = rev_num*10 + num%10; (b) Divide num by 10 (3) Return rev_num170Input: num (1) Initialize rev_num = 0 (2) Loop while num > 0 (a) Multiply rev_num by 10 and add remainder of num divide by 10 to rev_num rev_num = rev_num*10 + num%10; (b) Divide num by 10 (3) Return rev_num171Input: num (1) Initialize rev_num = 0 (2) Loop while num > 0 (a) Multiply rev_num by 10 and add remainder of num divide by 10 to rev_num rev_num = rev_num*10 + num%10; (b) Divide num by 10 (3) Return rev_num21

Reverse of no. is 26542____07

Reverse of no. is 26542____2178 Reverse of no. is 265452 Input: num (1) Initialize rev_num = 0 (2) Loop while num > 0 (a) Multiply rev_num by 10 and add remainder of num divide by 10 to rev_num rev_num = rev_num*10 + num%10; (b) Divide num by 10 (3) Return rev_num170Input: num (1) Initialize rev_num = 0 (2) Loop while num > 0 (a) Multiply rev_num by 10 and add remainder of num divide by 10 to rev_num rev_num = rev_num*10 + num%10; (b) Divide num by 10 (3) Return rev_num181Input: num (1) Initialize rev_num = 0 (2) Loop while num > 0 (a) Multiply rev_num by 10 and add remainder of num divide by 10 to rev_num rev_num = rev_num*10 + num%10; (b) Divide num by 10 (3) Return rev_num21

Reverse of no. is 26542____07

Reverse of no. is 26547

C++

Reverse of no. is 1234561

Reverse of no. is 1234562 Reverse of no. is 1234563 Reverse of no. is 1234564

Reverse of no. is 26547

Java

Reverse of no. is 1234563597

Reverse of no. is 26542____2205

Reverse of no. is 26542____2301

Reverse of no. is 26542____2212

Reverse of no. is 26542____2309

Reverse of no. is 26542____2216

Reverse of no. is 26542____2313

Reverse of no. is 26542____2222

Reverse of no. is 26542Reverse of no. is 26549 Reverse of no. is 12345635980

Reverse of no. is 26542______21 Input: num (1) Initialize rev_num = 0 (2) Loop while num > 0 (a) Multiply rev_num by 10 and add remainder of num divide by 10 to rev_num rev_num = rev_num*10 + num%10; (b) Divide num by 10 (3) Return rev_num50Input: num (1) Initialize rev_num = 0 (2) Loop while num > 0 (a) Multiply rev_num by 10 and add remainder of num divide by 10 to rev_num rev_num = rev_num*10 + num%10; (b) Divide num by 10 (3) Return rev_num51Input: num (1) Initialize rev_num = 0 (2) Loop while num > 0 (a) Multiply rev_num by 10 and add remainder of num divide by 10 to rev_num rev_num = rev_num*10 + num%10; (b) Divide num by 10 (3) Return rev_num16

Reverse of no. is 26542____254____246

Reverse of no. is 26547

Python3

Reverse of no. is 12345615 Reverse of no. is 12345616

C#

Reverse of no. is 26542____2125____2411

Reverse of no. is 26542____2413

Reverse of no. is 26542______09 Input: num (1) Initialize rev_num = 0 (2) Loop while num > 0 (a) Multiply rev_num by 10 and add remainder of num divide by 10 to rev_num rev_num = rev_num*10 + num%10; (b) Divide num by 10 (3) Return rev_num304 Input: num (1) Initialize rev_num = 0 (2) Loop while num > 0 (a) Multiply rev_num by 10 and add remainder of num divide by 10 to rev_num rev_num = rev_num*10 + num%10; (b) Divide num by 10 (3) Return rev_num347Input: num (1) Initialize rev_num = 0 (2) Loop while num > 0 (a) Multiply rev_num by 10 and add remainder of num divide by 10 to rev_num rev_num = rev_num*10 + num%10; (b) Divide num by 10 (3) Return rev_num418

Reverse of no. is 26542____2205

Reverse of no. is 26542____2347 Input: num (1) Initialize rev_num = 0 (2) Loop while num > 0 (a) Multiply rev_num by 10 and add remainder of num divide by 10 to rev_num rev_num = rev_num*10 + num%10; (b) Divide num by 10 (3) Return rev_num433

Reverse of no. is 26542____2212

Reverse of no. is 26542____2437

Reverse of no. is 26542____2216

Reverse of no. is 26542____250____21Input: num (1) Initialize rev_num = 0 (2) Loop while num > 0 (a) Multiply rev_num by 10 and add remainder of num divide by 10 to rev_num rev_num = rev_num*10 + num%10; (b) Divide num by 10 (3) Return rev_num443

Reverse of no. is 26542____2222

Reverse of no. is 26542Reverse of no. is 26549 Reverse of no. is 12345635980

Reverse of no. is 26542____21 Reverse of no. is 26548

Reverse of no. is 26542____044Input: num (1) Initialize rev_num = 0 (2) Loop while num > 0 (a) Multiply rev_num by 10 and add remainder of num divide by 10 to rev_num rev_num = rev_num*10 + num%10; (b) Divide num by 10 (3) Return rev_num46

Reverse of no. is 265446____16218

Reverse of no. is 26547

Javascript

Reverse of no. is 265402

Reverse of no. is 26542____2205

Reverse of no. is 26542____2482

Reverse of no. is 26542

Reverse of no. is 26542____2216

Reverse of no. is 26542____2493

Reverse of no. is 26542____2222

Reverse of no. is 265490

Reverse of no. is 265434

PHP

Reverse of no. is 265452

Reverse of no. is 26547

Reverse of no. is 265493 Input: num (1) Initialize rev_num = 0 (2) Loop while num > 0 (a) Multiply rev_num by 10 and add remainder of num divide by 10 to rev_num rev_num = rev_num*10 + num%10; (b) Divide num by 10 (3) Return rev_num46____2554Reverse of no. is 265457Reverse of no. is 265499

Reverse of no. is 265401

Đầu raReverse of no. is 123456

Độ phức tạp về thời gian. O(log10n) trong đó n là số đầu vào
Không gian phụ. Ô(1)

Các chữ số đảo ngược của một số nguyên có xử lý tràn

Lưu ý rằng chương trình trên không xem xét các số 0 đứng đầu. Ví dụ 100 chương trình sẽ in 1. Nếu bạn muốn in 001 thì hãy xem nhận xét này từ Maheshwar

Sử dụng Slicing trong Python

C++

Reverse of no. is 1234561

Reverse of no. is 1234562 Reverse of no. is 1234563 Reverse of no. is 1234564

Reverse of no. is 26547

Reverse of no. is 265490

Reverse of no. is 26547

Java

Reverse of no. is 1234566279 Reverse of no. is 1234566282

Reverse of no. is 1234563598 Reverse of no. is 1234563599

Reverse of no. is 26542____2205

Reverse of no. is 26542____2619

Reverse of no. is 26542____2212

Reverse of no. is 26542____2216

Reverse of no. is 26542____2629

Reverse of no. is 26542____2222

Reverse of no. is 26542Reverse of no. is 26549 Reverse of no. is 12345635980

Reverse of no. is 26542______21 Input: num (1) Initialize rev_num = 0 (2) Loop while num > 0 (a) Multiply rev_num by 10 and add remainder of num divide by 10 to rev_num rev_num = rev_num*10 + num%10; (b) Divide num by 10 (3) Return rev_num50Input: num (1) Initialize rev_num = 0 (2) Loop while num > 0 (a) Multiply rev_num by 10 and add remainder of num divide by 10 to rev_num rev_num = rev_num*10 + num%10; (b) Divide num by 10 (3) Return rev_num51Input: num (1) Initialize rev_num = 0 (2) Loop while num > 0 (a) Multiply rev_num by 10 and add remainder of num divide by 10 to rev_num rev_num = rev_num*10 + num%10; (b) Divide num by 10 (3) Return rev_num16

Reverse of no. is 26542____254____246

Reverse of no. is 26547

Python3

Reverse of no. is 12345615 Reverse of no. is 12345616

C#

Reverse of no. is 1234563598 Reverse of no. is 1234563599

Reverse of no. is 26542____2205

Reverse of no. is 26542____2212

Reverse of no. is 26542____2125____2731

Reverse of no. is 26542____2733

Reverse of no. is 26542____2623____2304 Input: num (1) Initialize rev_num = 0 (2) Loop while num > 0 (a) Multiply rev_num by 10 and add remainder of num divide by 10 to rev_num rev_num = rev_num*10 + num%10; (b) Divide num by 10 (3) Return rev_num347Input: num (1) Initialize rev_num = 0 (2) Loop while num > 0 (a) Multiply rev_num by 10 and add remainder of num divide by 10 to rev_num rev_num = rev_num*10 + num%10; (b) Divide num by 10 (3) Return rev_num738

Reverse of no. is 26542____2216

Reverse of no. is 26542____2742

Reverse of no. is 26542____2222

Reverse of no. is 26542Reverse of no. is 26549 Reverse of no. is 12345635980

Reverse of no. is 26542____21 Reverse of no. is 26548

Reverse of no. is 265446____16218

Reverse of no. is 26547

Javascript

Reverse of no. is 26547

Reverse of no. is 265490

Reverse of no. is 265404

Đầu raReverse of no. is 2654

Độ phức tạp về thời gian. O(n), trong đó n là số đầu vào
Không gian phụ. Ô(1)

Hãy thử các tiện ích mở rộng của các chức năng trên cũng sẽ hoạt động đối với các số dấu phẩy động

Phân chia và chinh phục

thuật toán

Reverse of no. is 12345662

Python3

Reverse of no. is 1234566279 Input: num (1) Initialize rev_num = 0 (2) Loop while num > 0 (a) Multiply rev_num by 10 and add remainder of num divide by 10 to rev_num rev_num = rev_num*10 + num%10; (b) Divide num by 10 (3) Return rev_num807

Reverse of no. is 265452Input: num (1) Initialize rev_num = 0 (2) Loop while num > 0 (a) Multiply rev_num by 10 and add remainder of num divide by 10 to rev_num rev_num = rev_num*10 + num%10; (b) Divide num by 10 (3) Return rev_num904Input: num (1) Initialize rev_num = 0 (2) Loop while num > 0 (a) Multiply rev_num by 10 and add remainder of num divide by 10 to rev_num rev_num = rev_num*10 + num%10; (b) Divide num by 10 (3) Return rev_num77 Input: num (1) Initialize rev_num = 0 (2) Loop while num > 0 (a) Multiply rev_num by 10 and add remainder of num divide by 10 to rev_num rev_num = rev_num*10 + num%10; (b) Divide num by 10 (3) Return rev_num829 Input: num (1) Initialize rev_num = 0 (2) Loop while num > 0 (a) Multiply rev_num by 10 and add remainder of num divide by 10 to rev_num rev_num = rev_num*10 + num%10; (b) Divide num by 10 (3) Return rev_num64Input: num (1) Initialize rev_num = 0 (2) Loop while num > 0 (a) Multiply rev_num by 10 and add remainder of num divide by 10 to rev_num rev_num = rev_num*10 + num%10; (b) Divide num by 10 (3) Return rev_num64 Input: num (1) Initialize rev_num = 0 (2) Loop while num > 0 (a) Multiply rev_num by 10 and add remainder of num divide by 10 to rev_num rev_num = rev_num*10 + num%10; (b) Divide num by 10 (3) Return rev_num15Input: num (1) Initialize rev_num = 0 (2) Loop while num > 0 (a) Multiply rev_num by 10 and add remainder of num divide by 10 to rev_num rev_num = rev_num*10 + num%10; (b) Divide num by 10 (3) Return rev_num72

Input: num (1) Initialize rev_num = 0 (2) Loop while num > 0 (a) Multiply rev_num by 10 and add remainder of num divide by 10 to rev_num rev_num = rev_num*10 + num%10; (b) Divide num by 10 (3) Return rev_num6Input: num (1) Initialize rev_num = 0 (2) Loop while num > 0 (a) Multiply rev_num by 10 and add remainder of num divide by 10 to rev_num rev_num = rev_num*10 + num%10; (b) Divide num by 10 (3) Return rev_num94Input: num (1) Initialize rev_num = 0 (2) Loop while num > 0 (a) Multiply rev_num by 10 and add remainder of num divide by 10 to rev_num rev_num = rev_num*10 + num%10; (b) Divide num by 10 (3) Return rev_num31Input: num (1) Initialize rev_num = 0 (2) Loop while num > 0 (a) Multiply rev_num by 10 and add remainder of num divide by 10 to rev_num rev_num = rev_num*10 + num%10; (b) Divide num by 10 (3) Return rev_num914Input: num (1) Initialize rev_num = 0 (2) Loop while num > 0 (a) Multiply rev_num by 10 and add remainder of num divide by 10 to rev_num rev_num = rev_num*10 + num%10; (b) Divide num by 10 (3) Return rev_num915Input: num (1) Initialize rev_num = 0 (2) Loop while num > 0 (a) Multiply rev_num by 10 and add remainder of num divide by 10 to rev_num rev_num = rev_num*10 + num%10; (b) Divide num by 10 (3) Return rev_num83Input: num (1) Initialize rev_num = 0 (2) Loop while num > 0 (a) Multiply rev_num by 10 and add remainder of num divide by 10 to rev_num rev_num = rev_num*10 + num%10; (b) Divide num by 10 (3) Return rev_num24Input: num (1) Initialize rev_num = 0 (2) Loop while num > 0 (a) Multiply rev_num by 10 and add remainder of num divide by 10 to rev_num rev_num = rev_num*10 + num%10; (b) Divide num by 10 (3) Return rev_num83Input: num (1) Initialize rev_num = 0 (2) Loop while num > 0 (a) Multiply rev_num by 10 and add remainder of num divide by 10 to rev_num rev_num = rev_num*10 + num%10; (b) Divide num by 10 (3) Return rev_num83Input: num (1) Initialize rev_num = 0 (2) Loop while num > 0 (a) Multiply rev_num by 10 and add remainder of num divide by 10 to rev_num rev_num = rev_num*10 + num%10; (b) Divide num by 10 (3) Return rev_num824Input: num (1) Initialize rev_num = 0 (2) Loop while num > 0 (a) Multiply rev_num by 10 and add remainder of num divide by 10 to rev_num rev_num = rev_num*10 + num%10; (b) Divide num by 10 (3) Return rev_num85Input: num (1) Initialize rev_num = 0 (2) Loop while num > 0 (a) Multiply rev_num by 10 and add remainder of num divide by 10 to rev_num rev_num = rev_num*10 + num%10; (b) Divide num by 10 (3) Return rev_num922

Input: num (1) Initialize rev_num = 0 (2) Loop while num > 0 (a) Multiply rev_num by 10 and add remainder of num divide by 10 to rev_num rev_num = rev_num*10 + num%10; (b) Divide num by 10 (3) Return rev_num6Input: num (1) Initialize rev_num = 0 (2) Loop while num > 0 (a) Multiply rev_num by 10 and add remainder of num divide by 10 to rev_num rev_num = rev_num*10 + num%10; (b) Divide num by 10 (3) Return rev_num94Input: num (1) Initialize rev_num = 0 (2) Loop while num > 0 (a) Multiply rev_num by 10 and add remainder of num divide by 10 to rev_num rev_num = rev_num*10 + num%10; (b) Divide num by 10 (3) Return rev_num31Input: num (1) Initialize rev_num = 0 (2) Loop while num > 0 (a) Multiply rev_num by 10 and add remainder of num divide by 10 to rev_num rev_num = rev_num*10 + num%10; (b) Divide num by 10 (3) Return rev_num914Input: num (1) Initialize rev_num = 0 (2) Loop while num > 0 (a) Multiply rev_num by 10 and add remainder of num divide by 10 to rev_num rev_num = rev_num*10 + num%10; (b) Divide num by 10 (3) Return rev_num929Input: num (1) Initialize rev_num = 0 (2) Loop while num > 0 (a) Multiply rev_num by 10 and add remainder of num divide by 10 to rev_num rev_num = rev_num*10 + num%10; (b) Divide num by 10 (3) Return rev_num83Input: num (1) Initialize rev_num = 0 (2) Loop while num > 0 (a) Multiply rev_num by 10 and add remainder of num divide by 10 to rev_num rev_num = rev_num*10 + num%10; (b) Divide num by 10 (3) Return rev_num24Input: num (1) Initialize rev_num = 0 (2) Loop while num > 0 (a) Multiply rev_num by 10 and add remainder of num divide by 10 to rev_num rev_num = rev_num*10 + num%10; (b) Divide num by 10 (3) Return rev_num83Input: num (1) Initialize rev_num = 0 (2) Loop while num > 0 (a) Multiply rev_num by 10 and add remainder of num divide by 10 to rev_num rev_num = rev_num*10 + num%10; (b) Divide num by 10 (3) Return rev_num83Input: num (1) Initialize rev_num = 0 (2) Loop while num > 0 (a) Multiply rev_num by 10 and add remainder of num divide by 10 to rev_num rev_num = rev_num*10 + num%10; (b) Divide num by 10 (3) Return rev_num824Input: num (1) Initialize rev_num = 0 (2) Loop while num > 0 (a) Multiply rev_num by 10 and add remainder of num divide by 10 to rev_num rev_num = rev_num*10 + num%10; (b) Divide num by 10 (3) Return rev_num85Input: num (1) Initialize rev_num = 0 (2) Loop while num > 0 (a) Multiply rev_num by 10 and add remainder of num divide by 10 to rev_num rev_num = rev_num*10 + num%10; (b) Divide num by 10 (3) Return rev_num922Input: num (1) Initialize rev_num = 0 (2) Loop while num > 0 (a) Multiply rev_num by 10 and add remainder of num divide by 10 to rev_num rev_num = rev_num*10 + num%10; (b) Divide num by 10 (3) Return rev_num83Input: num (1) Initialize rev_num = 0 (2) Loop while num > 0 (a) Multiply rev_num by 10 and add remainder of num divide by 10 to rev_num rev_num = rev_num*10 + num%10; (b) Divide num by 10 (3) Return rev_num24Input: num (1) Initialize rev_num = 0 (2) Loop while num > 0 (a) Multiply rev_num by 10 and add remainder of num divide by 10 to rev_num rev_num = rev_num*10 + num%10; (b) Divide num by 10 (3) Return rev_num85Input: num (1) Initialize rev_num = 0 (2) Loop while num > 0 (a) Multiply rev_num by 10 and add remainder of num divide by 10 to rev_num rev_num = rev_num*10 + num%10; (b) Divide num by 10 (3) Return rev_num940

Độ phức tạp về thời gian. O(log10(một nửa số chữ số trong số)), số biểu thị đầu vào đã cho.
Không gian phụ. Ô(1).

sử dụng chia và chinh phục

Sau đó, ta sẽ chia số đã cho thành hai phần, đảo ngược nửa đầu và nửa sau rồi nối chúng lại

Làm cách nào để đảo ngược một số trong C bằng chức năng?

Hiểu thuật toán của chương trình C để đảo ngược một số .

Chúng ta sẽ khai báo hai biến int. 'Number' lưu số cần đảo ngược và 'Reversed_Number', được khởi tạo là 0, sẽ giữ số bị đảo ngược kết quả

Bây giờ, chúng ta sẽ sử dụng công thức Reversed_Number = Reversed_Number*10 + Number%10

Làm thế nào để đảo ngược một giá trị trong C?

Đảo ngược một số nguyên . đảo ngược = đảo ngược * 10 + dư ; . vòng lặp kết thúc.

Làm thế nào để đảo ngược thứ tự của một số trong C?

Chương trình C đảo số .

int main() { int n, r = 0;

printf("Nhập số để đảo ngược\n");

trong khi (n. = 0) { r = r * 10;

printf("Số đảo ngược = %d\n", r);

Làm cách nào để đảo ngược một số có 3 chữ số trong C?

Cách đảo ngược số có 3 chữ số .

Khai báo int thành num1, num2 và num3 cùng với đầu ra, đảo ngược

Yêu cầu người dùng nhập số để đảo ngược

Lưu số bằng scanf

Lấy chữ số cuối cùng của số , num1= n/100

Lấy chữ số ở giữa từ người dùng, num2=(num%100)/10

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Làm cách nào để đảo ngược một số trong C bằng chức năng?

Làm thế nào để đảo ngược một giá trị trong C?

Làm thế nào để đảo ngược thứ tự của một số trong C?

Làm cách nào để đảo ngược một số có 3 chữ số trong C?

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