7 Reverse Integer

链接：https://leetcode.com/problems/reverse-integer/
问题描述：
Reverse digits of an integer.

Example1: x = 123, return 321
Example2: x = -123, return -321

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Have you thought about this?
Here are some good questions to ask before coding. Bonus points for you if you have already thought through this!

If the integer’s last digit is 0, what should the output be? ie, cases such as 10, 100.

Did you notice that the reversed integer might overflow? Assume the input is a 32-bit integer, then the reverse of 1000000003 overflows. How should you handle such cases?

For the purpose of this problem, assume that your function returns 0 when the reversed integer overflows.

Update (2014-11-10):
Test cases had been added to test the overflow behavior.

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这个问题就是求数字的倒序，需要注意的问题如下：
1.负数需要保持符号。
2.需要注意溢出的问题。一个int可以表示数字范围是2147483647— -2147483648。当逆序的数字超过这个范围，那么函数返回0。

class Solution {public:    long long int reverse(int x) {        bool flag=true;        vector<int> vi;        long long int result=0;        if(x<0)        {            flag=false;            x=-x;        }           while(x)        {            vi.push_back(x%10);            x/=10;        }        for(int i=0;i<vi.size();i++)            result+=vi[i]*pow(10,vi.size()-1-i);        if(!flag)           result= -result;        if(result> INT_MAX || result<INT_MIN)            return 0;        else            return result;    }};

稍微改进下代码：

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