Leetcode 7 Reverse Integer 倒序整数

时间 2020-01-29

标签 leetcode reverse integer 倒序整数繁體版

原文原文链接

原题目为:java

Reverse digits of an integer.
Example1: x = 123, return 321

Example2: x = -123, return -321git

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!this

If the integer's last digit is 0, what should the output be? ie, cases
such as 10, 100.code

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?input

For the purpose of this problem, assume that your function returns 0
when the reversed integer overflows.it

难度: Easyio

此题让咱们输出给定一个整数的倒序数, 好比123倒序为321, -123倒序为-321. 可是若是倒序的过程当中发生整型溢出, 咱们就输出0.ast

倒序不复杂, 关键在于如何断定将要溢出.function

最终AC的程序以下:class

public class Solution {

    public int reverse(int x) {
        int x1 = Math.abs(x);
        int rev = 0;
        while (x1 > 0) {
            if (rev > (Integer.MAX_VALUE - (x1 - (x1 / 10) * 10)) / 10) {
                return 0;
            }
            rev = rev * 10 + (x1 - (x1 / 10) * 10);
            x1 = x1 / 10;
        }
        if (x > 0) {
            return rev;
        } else {
            return -rev;
        }
    }
}

其中 x1 - (x1 / 10) * 10 是获取x1的个位数字, 断定下一步是否将要溢出, 使用 rev > (Integer.MAX_VALUE - (x1 - (x1 / 10) * 10)) / 10 .