标签：str bre style etc log eve handle sed public
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
题意：
给定一个10进制整数，翻转它。
Solution1: directly do the simulation.
Two tricky parts to be handled:
(1) overflow : 32-bit signed integer range: [−231, 231 − 1], whihc means [−2,147,483,648 2,147,483,647]. What if input is 2,147,483,647, after reversing, it will be 7,463,847,412.
(2) negative numbers
code:
1 /* 2 Time Complexity: O(log(n)) coz we just travese half part of original input 3 Space Complexity: O(1) 4 */ 5 class Solution { 6 public int reverse(int input) { 7 long sum = 0; 8 while(input !=0){ 9 sum = sum*10 + input %10; 10 input = input /10; 11 12 if(sum > Integer.MAX_VALUE || sum < Integer.MIN_VALUE){ 13 sum = 0; // returns 0 when the reversed integer overflows 14 break; 15 } 16 } 17 return (int)sum; 18 } 19 }
[leetcode]7. Reverse Integer反转整数
标签：str bre style etc log eve handle sed public
原文地址：https://www.cnblogs.com/liuliu5151/p/10652711.html
