Single Number I

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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?

分析：两个相同的数进行异或，结果为0。对所有数据进行异或，则可以得到所有数字中只出现一次的那个。

运行时间：18ms

 1 class Solution {
 2 public:
 3     int singleNumber(vector<int>& nums) {
 4         if(nums.size() == 0) return 0;
 5         
 6         int result = 0;
 7         for(int i = 0; i < nums.size(); i++) result ^= nums[i];
 8         
 9         return result;
10     }
11 };

Single Number I

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原文地址：http://www.cnblogs.com/amazingzoe/p/4508737.html