Reverse Integer

Reverse Integer

 

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).

对于边界检测是很重要的一点，也是基础问题。

int 的范围为 -2147483648— 2147483647 ，所以在反转时要考虑溢出问题。

在反转最后一位前，应判断生成的数字与 （最大值 - |数值| % 10）/ 10 的大小，如果大于的话，生成的数字在加上最后一位后必然大于最大值，会溢出。

public class Solution {    public int reverse(int x) {        if (x == Integer.MIN_VALUE)             return Integer.MIN_VALUE;        int result = 0;        int num = Math.abs(x);        while (num != 0) {            if (result > (Integer.MAX_VALUE - num % 10) / 10)                return x > 0 ? Integer.MAX_VALUE : Integer.MIN_VALUE;            result = result * 10 + num % 10;            num /= 10;        }        return x < 0 ? -result : result;     }}

Reverse Integer

Submission Details

1020 / 1020 test cases passed.

Status: 

Accepted

Runtime: 476 ms

Submitted: 0 minutes ago