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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?
class Solution { public: int singleNumber(int A[], int n) { int result = A[0]; int i = 1; while(i<n) { result = A[i] ^ result; //异或，相同为0 不同为1 i++; } return result; } };
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原文地址：http://www.cnblogs.com/qionglouyuyu/p/5022844.html
