【LeetCode】7. Reverse Integer问题解析

Reverse digits of an integer.
Example1: x = 123, return 321
Example2: x = -123, return -321
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Have you thought about this?
Here are some good questions to ask before coding. Bonuspoints for you if you have already thought through this!
If the integer's last digit is 0, what should the outputbe? ie, cases such as 10, 100.
Did you notice that the reversed integer might overflow?Assume the input is a 32-bit integer, then the reverse of 1000000003 overflows.How should you handle such cases?
For the purpose of this problem, assume that yourfunction returns 0 when the reversed integer overflows.
 
Note:

The input is assumed to be a 32-bit signed integer. Your function should return0 when the reversed integer overflows.
//本题是一道数字翻转的问题，问题不难，但是最容易忽略的一点就是数据的溢出问题！

Int类型的数据范围是： -2147483648～2147483647 
但是如果将2147483647这个数进行翻转，很明显，溢出！

所以这里使用long long类型的数据储存！

用到的一个核心就是

long long L=0;

L=L*10+x%10;

最终代码！核心就是/10与%10

class Solution {

public:

   int reverse(int x) {

       //int类型范围28亿  可能会存在溢出的问题！so

       long long y=0;

       while(x!=0){

           y=10*y+x%10;

           x=x/10;

       }

     return (y > INT_MAX || y < INT_MIN) ? 0 : y;

    }

};