Leetcode 136. 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?

题目要求O(n)时间复杂度，O(1)空间复杂度。

思路：利用异或操作。异或的性质1：交换律a ^ b = b ^ a，性质2：0 ^ a = a。于是利用交换律可以将数组假想成相同元素全部相邻，于是将所有元素依次做异或操作，相同元素异或为0，最终剩下的元素就为Single Number。时间复杂度O(n)，空间复杂度O(1)

C++:

class Solution {
public:
    int singleNumber(vector<int>& nums) {
        int len = nums.size(), a = 0;
        for(int i = 0; i < len; i++){
            a ^= nums[i];
        }
        return a;
    }
};

 

c:

int singleNumber(int* nums, int numsSize) {
    int i, a = 0;
    for(i = 0; i < numsSize; i++){
        a ^= nums[i];
    }
    return a;
}