大纲
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题目
地址
内容
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
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.
解题
这题主要考察“异或”操作。
异或操作具有以下几个性质：
- 自反性：x ^ y ^ y = x
- 结合性：(x ^ y) ^ z = x ^ (y ^ z)
- 与数字0异或无变化：x ^ 0 = x
因为x ^ x =0，这样偶数个x就会得到结果0；而x ^ 0 = x，这样会让单个的x得以保留。所以这题的思路就是对所有数字进行异或操作。
#include <vector>
using namespace std;class Solution {
public:int singleNumber(vector<int>& nums) {int result = 0;for (int num : nums) {result ^= num;}return result;}
};
代码地址
https://github.com/f304646673/leetcode/tree/main/136-Single-Number/
