leetcode题解 7.Reverse Integer

题目：

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.

这个题目是一个easy的题目，但是细节要注意，第一点要注意记得处理负数的情况，

第二点就是记得看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.

就是超过int的范围了就需要返回0。这一点非常重要，因为你要判断的是可能反转后会溢出，那么需要注意的就是

需要用一个long long来存储答案ans。判断使用的语句就是：

if(ans>INT_MAX||ans<INT_MIN)
            return 0;

啦啦啦！

代码如下：

class Solution {
public:
    int reverse(int x) {
        long long ans=0;
        bool isLittle=false;
        if(x<0)
        {
            x=-x;
            isLittle=true;
        }
        while(x>0)
        {
            int num=x%10;
            ans=ans*10+num;
            x=x/10;
        }
        if(ans>INT_MAX||ans<INT_MIN)
            return 0;
        if(isLittle)
            return -ans;
        else return ans;
    }
};