[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: for each iteration, we do multiplication or division with a position number -- 10 , which means if sign is '-' , the sign will be kept all the time.

 

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 return 0; // returns 0 when the reversed integer overflows 14 } 15 } 16 return (int)sum; 17 } 18 }

 

转载于:https://www.cnblogs.com/liuliu5151/p/10652711.html