【LeetCode】7. Reverse Integer

对于a,b,取余和取模运算分为两步：

（1）求整数商：

（2）计算模或者余数：

取余和取模的差别在于求整数商的不同：取余，遵循尽可能让商向0靠近的原则；取模，遵循尽可能让商向负无穷靠近的原则。

e.g.,

4rem(-3)=1(商取-1) 4mod(-3)=-2(商取-2)

简单地，对于负数对正数的%运算，可以用绝对值计算后再直接计算符号。

e.g.,

cout<<4%(-3)<<" "<<-4%3<

LeetCode 7. Reverse Integer

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

Example 1:

Input: 123
Output: 321

Example 2:

Input: -123
Output: -321

Example 3:

Input: 120
Output: 21

Note:
Assume we are dealing with an environment which could only store integers within the 32-bit signed integer range: [−231,  231 − 1]. For the purpose of this problem, assume that your function returns 0 when the reversed integer overflows.

class Solution {
public:
    int reverse(int x) {
        long long int ans=0; 
        while(x){
            ans=ans*10+x%10;
            x/=10;
        }
        if(ans>INT_MAX||ans