LeetCodeOJ——7. Reverse Integer

Reverse Integer

Reverse digits of an integer.

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

**spoilers：
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.**

解题思路：题目本身并不难，难的是考虑边界条件，最重要的越界。这里注意两点，用16进制数表示int的最大值最小值更加容易， int intMax=0x7fffffff; int intMin=0x80000000; 此外用long来防止越界，但能判断int是否越界。

代码：

class Solution {public:    int reverse(int x) {        int flag=1;        if(x<0){            flag=0;            x=-x;        }        queue<int> q;        while(x>0){            q.push(x%10);            x=x/10;        }        int intMax=0x7fffffff;  //can use 0x????        int intMin=0x80000000;        long sum=0;  // 可以 可以用 可以用long        int temp=0;        while(!q.empty()){            temp=q.front();            q.pop();            sum=sum*10+temp;        }        if(flag==0){            sum=-sum;        }        if(sum>intMax||sum<intMin){            return 0;        }        return (int)sum;    }};