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：用下列multiset时间复杂度O（n），空间复杂度O（n），因为额外用的multiset的大小和数组大小相同。
#include<set> using namespace std; class Solution { public: int singleNumber(int A[], int n) { multiset<int> ms; for(int i=0;i<n;i++) ms.insert(A[i]); for(int i = 0;i<n;i++) { int num = ms.count(A[i]); if(num == 1) return A[i]; } } };
分析：用“异或的方法”，原理是：两个相同的数异或等于0，0和任意数A异或等于A。
class Solution { public: int singleNumber(int A[], int n) { int a=0; for(int i=0;i<n;i++) a = a^A[i]; return a; } };
[LeetCode] Single Number,布布扣,bubuko.com
原文：http://www.cnblogs.com/Xylophone/p/3779237.html