LeetCode Problem 136:Single Number

描述：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?

題目要求O(n)時間復雜度，O(1)空間復雜度。

思路1：初步使用暴力搜索，遍歷數組，發現兩個元素相等，則將這兩個元素的標志位置為1，最后返回標志位為0的元素即可。時間復雜度O(n^2)沒有AC，Status:Time Limit Exceed

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

思路2：利用異或操作。異或的性質1：交換律a ^ b = b ^ a，性質2：0 ^ a = a。於是利用交換律可以將數組假想成相同元素全部相鄰，於是將所有元素依次做異或操作，相同元素異或為0，最終剩下的元素就為Single Number。時間復雜度O(n)，空間復雜度O(1)

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