Reverse Integer--LeetCode

Reverse Integer

（原题链接：点击打开链接）

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.

Solution：

把一个整数反过来显示，最简单的想法就是不断地取到这个整数的最后一位，然后结果左移一位，该整数向右移一位，直到该整数变为0。

由于这个数为32位有符号整数，所以要判断溢出

class Solution {public:    int reverse(int x) {        long long int result = 0;        int fuhao = x < 0 ? 1 : 0;        x = x < 0 ? -x : x;        while(x)        {            result = result * 10 + x % 10;            x = x / 10;            //cout<<result<<endl;        }        if (fuhao) result = -result;        if (result < INT32_MIN || result > INT32_MAX) return 0;        return result;    }};

我们注意到，其实对于符号的判断是不必要的。因为取余的结果和被除数是一样的，所以我们可以对算法进行些微的改进。可是时间复杂度依然为O（N），N为x的长度

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