[LeetCode 7] Reverse Integer

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[Problem Link]

Description

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

Example

Input: 123
Output: 321

Input: -123
Output: -321

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.

Solution

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

Summary

• 网上判断数值是否溢出有很多不同的方法，经对比我觉得这种最好理解，所以选择了这一种。INT型占8个字节，32位。long型占16个字节，64位。用long型计算出数值，判断是否在[INT_MIN, INT_MAX]范围内。