leetcode week16

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.

Note:

The input is assumed to be a 32-bit signed integer. Your function should return 0 when the reversed integer overflows.

问题描述：将一个整数反转，比如123变成321，-123变成-321这样。

解题思路：要考虑的地方，首先是如果出现末尾为0的情况要如何处理，比如100，反转之后应该是要变成1；

  然后是要注意数据溢出的情况，比如说输入是32位整数，1000000003反转后将溢出，此时返回0；

考虑完这两个问题之后，写算法的时候就是先记录最后一位（用mod的方法）然后累加最后一位乘10这样。

具体代码如下：

class Solution {public:    int reverse(int x) {        int digit = 0;        int carry = 0;        while (x != 0) {            carry = x % 10;            if (digit > INT32_MAX / 10 || digit < INT32_MIN / 10){                return 0;            }            digit = digit * 10 + carry;            x /= 10;        }        return digit;    }};