Reverse Integer——LeetCode进阶路⑦

原题链接https://leetcode.com/problems/reverse-integer/

• 题目描述

  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：从高位开始对10求余，存下来乘10，依次进行……这道题实在是太温柔~
  唯一需要注意的不能用粗暴法，用数组存起来再反转，会溢出的

• 源码附录:
   

  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;
      }
  }