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/*
* @lc app=leetcode.cn id=7 lang=cpp
*
* [7] 整数反转
*
* https://leetcode-cn.com/problems/reverse-integer/description/
*
* algorithms
* Easy (31.59%)
* Total Accepted: 106.3K
* Total Submissions: 330.1K
* Testcase Example: '123'
*
* 给出一个 32 位的有符号整数,你需要将这个整数中每位上的数字进行反转。
*
* 示例 1:
*
* 输入: 123
* 输出: 321
*
*
* 示例 2:
*
* 输入: -123
* 输出: -321
*
*
* 示例 3:
*
* 输入: 120
* 输出: 21
*
*
* 注意:
*
* 假设我们的环境只能存储得下 32 位的有符号整数,则其数值范围为 [−2^31,  2^31 − 1]。请根据这个假设,如果反转后整数溢出那么就返回
* 0。
*
*/
//stack + overflow
class Solution {
public:
int reverse(int x) {
int ret = 0;
while (x)
{
//pop
int pop = x % 10;
x /= 10;
//positive overflow
if (ret > INT_MAX/10 || ret == INT_MAX/10 && pop > 7) return 0;
//negative overflow
if (ret < INT_MIN/10 || ret == INT_MIN/10 && pop > 8) return 0;
//push
ret = ret * 10 + pop;
}
return ret;
}
};
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