[LeetCode] Reverse Integer [8]

题目

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?

Throw an exception? Good, but what if throwing an exception is not an option? You would then have to re-design the function (ie, add an extra parameter).

原题地址

解题思路

颠倒一个整数中数字的位置，哈哈，类似翻转字符串一样，只不过整数需要计算，需要考虑溢出，其他无而。写出代码不难，只是溢出的时候怎么办，我这里是按照返回-1处理的，在leetcode上可以AC。

代码实现：

class Solution {
public:
    int reverse(int x) {
        int sign = 1;
        if(x < 0) sign = -1;
        int ret =0;
        while(x!=0){
            ret = ret*10 + abs(x%10)*sign;
            x /= 10;
        }
        if((x>0 && ret <0) && (x<0 && ret>0)) return -1;
        return ret;
    }
};

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