Single Number

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Given an array of integers, every element appears twice except for one. Find that single one.

Note:
Your algorithm should have a linear runtime complexity. Could you implement it without using extra memory?

思路：a^b^b=a a抑或同一个数两次认为a。

 1 class Solution {
 2 public:
 3     int singleNumber(int A[], int n) {
 4         int res;
 5         for(int i=0;i<n;i++)
 6         {
 7             res=res^A[i];
 8         }
 9         return res;
10     }
11 };

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Single Number

标签：style   blog   color   strong   for   io   

原文地址：http://www.cnblogs.com/hicandyman/p/3825802.html