Single Number

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136.Single Number

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

You must implement a solution with a linear runtime complexity and use only constant extra space.

Example 1:

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

Example 2:

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

Example 3:

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Input: nums = [1]
Output: 1

时间复杂度：O(n)

空间复杂度：O(1)

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class Solution {
public:
    int singleNumber(vector<int>& nums) {
        sort(nums.begin(), nums.end());
        for (vector<int>::iterator it = nums.begin(); it != nums.end(); it++) {
            int temp = *it;
            it++;
            if (it == nums.end()||temp != *it)return temp;
        }
        return 0;
    }
};

位运算：

时间复杂度：O(n)

空间复杂度：O(1)

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class Solution {
    public:
    int singleNumber(vector<int>& nums) {
        int ret = 0;
        for (auto e: nums) ret ^= e;
        return ret;
    }
};