## 题目

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 Example 3:

Input: nums =  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.

## 题解

1. 位运算

XOR 运算满足交换律和结合律：a⊕b⊕a = (a⊕a)⊕b = 0⊕b = b

 ``````1 2 3 4 5 6 7 8 9 `````` ``````class Solution { public int singleNumber(int[] nums) { int single=0; for (int num : nums) { single = single ^ num; } return single; } } ``````
1. 哈希表
 `````` 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 `````` ``````class Solution { public int singleNumber(int[] nums) { HashMap hs = new HashMap<>(); for (int num : nums) { hs.put(num, hs.getOrDefault(num, 0) + 1); } int single = 0; for (Map.Entry entry : hs.entrySet()) { if (entry.getValue() == 1) { single = entry.getKey(); } } return single; } } ``````