7. Reverse Integer

最新推荐文章于 2022-04-01 15:57:39 发布

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本文介绍了一个算法挑战：在32位有符号整数范围内翻转整数的每一位。通过实例展示了输入123返回321，输入-123返回-321的过程。特别注意，算法需判断翻转后的整数是否超出范围[-2^31, 2^31-1]，并在溢出时返回0。

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.

翻转字符串，简单但是有陷阱，需要判断翻转后是否超过了int范围，范围在[-2147483648,2147483647]，需要中间判断一下，乘以10后是否越界

    static class Solution {
        public int reverse(int x) {
            if (x == 0) return x;
            boolean mines = false;
            if (x < 0){
                mines = true;
                x = Math.abs(x);
            }
            int ans = 0;
            while (x > 0)
            {
                if (ans > Integer.MAX_VALUE / 10) return 0;
                ans = ans * 10 + x % 10;
                x = x / 10;
            }
            if (mines)
                ans = -ans;
            return ans;
        }
    }