leetcode 7. Reverse Integer

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于 2017-11-18 18:24:32 发布

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本文介绍了一种方法，用于反转32位带符号整数的数字，并通过一个C++实现示例展示了如何处理正负整数的反转，同时考虑了整数溢出的情况。

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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 hold integers within the 32-bit signed integer range. For the purpose of this problem, assume that your function returns 0 when the reversed integer overflows.

class Solution {
public:
    int reverse(int x)
    {   
        if(x==0 || x==INT_MIN)
            return 0;
        
        bool fushu = false;
        if (x < 0)
        {
            fushu = true; 
            x = -x;
        }
        vector<int> m1;
        while (x != 0)
        {
            m1.push_back(x % 10);
            x = x / 10;
        }
            
        long long sum = m1[0];
        int k = 1;
        while(k < m1.size())
        {
            sum = sum * 10 + m1[k];
            k++;
        }
        
        if (sum > 2147483648)
            return 0;
            
        return fushu ? -sum : sum;
    }
};