题目描述

题目链接：7. Reverse Integer

Given a 32-bit signed integer, reverse digits of an integer.

例子

例子 1

Input: 123
Output: 321

例子 2

Input: -123
Output: -321

例子 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.

解题思路

题目比较简单，只需要不断求余数即可，注意由于正负数转化的问题我们需要先将 x 转为长整形避免溢出，对结果变量也是一样。代码如下：

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#include <cmath>

class Solution {
public:
    int reverse(int x) {
        long rev = 0;
        long copy = x;
        bool is_negative = copy < 0;
        copy = copy > 0 ? copy : -copy;
        while (copy) {
            rev *= 10;
            rev += (copy % 10);
            copy /= 10;
        }
        if (is_negative) rev = -rev;
        if (rev < -1 * pow(2, 31) || rev > pow(2, 31) - 1) return 0;
        return rev;
    }
};

• 时间复杂度: O(digits of x)
• 空间复杂度: O(1)

GitHub 代码同步地址： 7.ReverseInteger.cpp

其他题目：
GitHub: Leetcode-C++-Solution
博客： Leetcode-Solutions