136. Single Number

Given a non-empty array of integers nums, every element appears twice except for one. Find that single one.

Follow up: Could you implement a solution with a linear runtime complexity and without using extra memory?

Example 1:

Input: nums = [2,2,1]
Output: 1

Example 2:

Input: nums = [4,1,2,1,2]
Output: 4

Constraints:

• 1 <= nums.length <= 3 * 104
• -3 * 104 <= nums[i] <= 3 * 104
• Each element in the array appears twice except for one element which appears only once.

解法一：遍历数组中的每个数，如果这个数存在集合s中，就从集合中删掉此数，否则，把这个数插入集合中。遍历结束后，集合中剩下的唯一的数就是single number。

时间复杂度O(n),空间复杂度O(n)

class Solution {
public:
    int singleNumber(vector<int>& nums) {
        unordered_set<int> s;
        int n=nums.size();
        for(int i=0;i<n;i++){
            if(s.count(nums[i]))
                s.erase(nums[i]);
            else s.insert(nums[i]);
        }
        return *s.begin();
    }
};

Runtime: 32 ms, faster than 20.54% of C++ online submissions for Single Number.

Memory Usage: 19.9 MB, less than 19.77% of C++ online submissions for Single Number.

解法二、用异或运算。一个数与其本身进行异或，等于0，一个数和0进行异或，等于这个数本身。异或运算满足交换律和结合律，即A^B^A=B。

时间复杂度O(n),空间复杂度O(1)

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

Runtime: 12 ms, faster than 97.15% of C++ online submissions for Single Number.

Memory Usage: 16.8 MB, less than 97.90% of C++ online submissions for Single Number.