1. 程式人生 > >Reverse Integer——LeetCode進階路⑦

Reverse Integer——LeetCode進階路⑦

原題連結https://leetcode.com/problems/reverse-integer/

  • 題目描述

    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.

  • 思路分析:從高位開始對10求餘,存下來乘10,依次進行……這道題實在是太溫柔~
    唯一需要注意的不能用粗暴法,用陣列存起來再反轉,會溢位的

  • 原始碼附錄:
     

    class Solution {
        public int reverse(int x) {
            if((x<=0 && x>-10)||(x<10&&x>0)){
                return x;
            }
            
            long result = 0;
            while(x != 0 ){
                result = result*10 + x%10;
                x = x/10;
            }
            
            if(result <= Integer.MIN_VALUE || result >= Integer.MAX_VALUE){
                return 0;
            }
            
            return (int)result;
        }
    }