[Easy] 7. Reverse Integer

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.

Solution

class Solution {
public:
    int reverse(int x) {
        int val=0;
        while(x!=0)
        {
            int pop = x%10;
            x /= 10;
            if( val > INT_MAX/10 || (val == INT_MAX/10 && pop > 7) ) return 0;
            if( val < INT_MIN/10 || (val == INT_MIN/10 && pop < -8)) return 0;
            val = val*10 + pop;
        }
        return val;
    }
};

解题思路

这道题考察点主要在 字符串反转后整型数是否溢出，需要判断是否溢出。