LeetCode 7. Reverse Integer C++

最新推荐文章于 2021-01-23 15:29:33 发布

原创最新推荐文章于 2021-01-23 15:29:33 发布 · 158 阅读

0 ·

CC 4.0 BY-SA版权

C++ 同时被 2 个专栏收录

26 篇文章

订阅专栏

LeetCode

13 篇文章

订阅专栏

本文探讨了在32位有符号整数范围内反转整数的方法。提供了C++实现的代码示例，包括溢出检查逻辑，确保反转后的整数仍处于有效范围内。

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: [−2^31, 2^31 − 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) {
        int res=0;
        while(x){
            int pop=x%10;            
            if(res>INT_MAX/10 || res<INT_MIN/10)
                return 0;
            res=res*10+pop;
            x/=10;
        }    	
	    return res;        
    }
};

神奇地通过了，感觉循环最后要加个判断：if (rev == INT_MAX / 10 && pop > 7)、if(rev == INT_MIN / 10 && pop < -8)

还有一个思路：