LeetCode—reverse-integer（反转整数）—java

题目描述：

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).

题意：反转，注意正负号，注意越界问题。

思路解析：

• 这个题反转主要是使用求余数和除法
• 但是注意最小值和最大值的情况
• 使用Math.abs()求绝对值反转数字

代码：

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