7. Reverse Integer

官方的鏈接：7. Reverse Integer

Description ：

Given a 32-bit signed integer, reverse digits of an integer.

Example1:

Input: 123

Output: 321

Example2:

Input: -123

Output: -321

Example3:

Input: 120

Output: 21

Note:

Assume we are dealing with an environment which could only hold integers within the 32-bit signed integer range. For the purpose of this problem, assume that your function returns 0 when the reversed integer overflows.

問題描述

反轉32位的數字

思路

上一次的模*10 + 這一次的模。中間判斷是否有溢出

註意：last_mod * 10 + this_mod，而x /= 10

[github-here]

 1 public class Q7_ReverseInteger {
 2     public int reverse(int x) {
 3         int revResult = 0;
 4         while (0 != x){
 5             int newResult = revResult * 10 + x % 10;
 6             //judge whether it overflows
 

 7             if (newResult / 10 != revResult) {
 8                 return 0;
 9             }
10             revResult = newResult;
11             x /= 10;
12         }
13         return revResult;
14     }
15 }

Q7：Reverse Integer