7. Reverse Integer - Easy

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.

判断是否溢出：rev是上一轮累加的结果，先把当前值减去个位数，再除以10，和之前的rev比较是否相同，如果不同说明精度丢失、溢出了，返回0；如果相同，更新rev，进入下一轮循环

time: O(log_10(x)), space: O(1)

class Solution {
    public int reverse(int x) {
        int rev = 0;
        while(x != 0) {
            int tmp1 = x % 10;
            int tmp2 = rev * 10 + tmp1;
            if((tmp2 - tmp1) / 10 != rev)
                return 0;
            rev = tmp2;
            x /= 10;
        }
        return rev;
    }
}

7. Reverse Integer - Easy

原文：https://www.cnblogs.com/fatttcat/p/10134950.html