Leetcode--Single Number

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?

Solution1:

时间复杂度O(n) 空间复杂度O(n)

class Solution {public:    int singleNumber(int A[], int n) {        map<int,int> m;        if(n<=0)            return 0;        else if(n==1)            return A[0];        for(int i=0;i<n;i++)            m[A[i]]++;                    for(int i=0;i<n;i++)        {            if(m[A[i]]==1)                return A[i];        }    }};

Solution2:

考虑数的性质，两个相同的数进行异或运算得到的结果是0

a^a=0;

0^a=a;

class Solution {public:    int singleNumber(int A[], int n) {        int res=0;        for(int i=0;i<n;i++)            res=res^A[i];                    return res;    }};