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Leetcode

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:

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

Example 2:

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

Example 3:

Input: nums = [1]
Output: 1

对所有元素异或，相同的元素异或得到 0 ，最后的结果即单个元素

class Solution {
    public int singleNumber(int[] nums) {
        int ret=0;
        for(int i:nums){
            ret = ret ^ i;
        }
        return ret;
    }
}

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

class Solution:
    def singleNumber(self, nums: List[int]) -> int:
        ret=0
        for i in nums:
            ret ^= i
        return ret