Reverse Integer——LeetCode進階路⑦
阿新 • • 發佈:2018-12-10
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; } }