【LeeCode】Reverse Integer 总结

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 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.

难点：如何判断int是否溢出？
总结：加上某值后再减去某值 与原值做对比 如果大小不变则没有溢出！

这是我的代码

class Solution {public:    int reverse(int x) {        int ret=0;        while(x!=0)        {            //如果大小改变则溢出，则返回0；            if(ret!=ret*10/10)            {                return 0;            }            ret*=10;            //同上            if(ret!=ret+x%10-x%10)            {                return 0;            }            ret+=x%10;            x=(x-x%10)/10;        }        return ret;    }};

这个是效率最高的代码：
还没有看，先放这里。

static int x = []() {     std::ios::sync_with_stdio(false);     cin.tie(NULL);      return 0; }();class Solution {public:    int reverse(int x) {        long answer = 0;        while (x != 0) {            answer = answer * 10 + x % 10;            if (answer > INT_MAX || answer < INT_MIN) return 0;            x /= 10;        }        return (int)answer;    }};