[leetcode]7-Reverse Integer

7. Reverse Integer

1)题目

Given a 32-bit signed integer, reverse digits of an integer.

Example 1:

Input: 123
Output: 321
Example 2:

Input: -123
Output: -321
Example 3:

Input: 120
Output: 21
Note:
Assume we are dealing with an environment which could only store integers within the 32-bit signed integer range: [?231,  231 ? 1]. For the purpose of this problem, assume that your function returns 0 when the reversed integer overflows.

2）思路

很简单,直接翻转加上临界溢出判断

3) 代码

public static  int reverse(int x) {
        int res = 0;
        while (x != 0) {
            if (Math.abs(res) > Integer.MAX_VALUE/10) {
                return 0;
            }
            res = res * 10 + x % 10;
            x/=10;
        }
        return res;
    }

4) 结果

时间复杂度:O(n)
空间复杂度:O(1)
耗时:

5) 调优

[leetcode]7-Reverse Integer

原文：https://www.cnblogs.com/novaCN/p/10328052.html