[leetcode]7. Reverse Integer反转整数

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 }

原文地址：https://www.cnblogs.com/liuliu5151/p/10652711.html