GuilinDev

Lc0007

05 August 2008

7 - Reverse Integer

原题概述

Given a 32-bit signed integer, reverse digits of an integer.

Example 1:

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Input: 123
Output: 321

Example 2:

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Input: -123
Output: -321

Example 3:

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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.

题意和分析

将一个无符号的正整数颠倒顺序，注意32位是整数的范围，如果超出则返回0。通过取模得到尾数，拿到后在新数中每次乘以10，而原来的数每次除以10消除掉尾数，判断是否overflow就减去尾数再除以10，看是否等于之前的result，如果不是就表明overflow了，返回0，否则正常计算。

时间O(n)，空间O(1)。

代码

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class Solution {
    public int reverse(int x) {
        int result = 0;
        while (x != 0) {
            int tail = x % 10;
            int newResult = result * 10 + tail; //判断是否溢出，把之前的操作反过来做一遍即可
            if ((newResult - tail) / 10 != result) {
                return 0;
            }
            result = newResult;
            x /= 10;
        }
        return result;
    }
}

正常做法

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class Solution {
   public int reverse(int x) {
      long result = 0;
      while (x != 0) {
         result = (result * 10) + (x % 10); // 不用单独判断正负号
         if (result > Integer.MAX_VALUE || result < Integer.MIN_VALUE) {
            return 0;
         }
         x /= 10;
      }
      return (int)result;
   }
}
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