XOR Operation Classical Algorithm Problem--Numbers that only appear once

Any number is XORed, and the result is still the original number, that is, a⊕0=a.

Any number is XORed with itself, and the result is 0, that is, a⊕a=0.

XOR operation satisfies commutative law and associative law, that is, a⊕b⊕a=b⊕a⊕a=b⊕(a⊕a)=b⊕0=b.

136. Numbers that occur only once

Given a non-empty array of integers, each element appears twice except for a certain element that appears only once. Find the element that appears only once.
Explanation:
Your algorithm should have linear time complexity. Can you do it without using extra space?
Example 1:
Input: [2,2,1]
Output: 1
Example 2:
Input: [4,1,2,1,2]
Output: 4

class Solution {
    
    
    public int singleNumber(int[] nums) {
    
    
        int single = 0;
        for (int num : nums) {
    
    
            // 把所有的数组的所有元素异或起来，两两相同的都异或为0，
            // 那么结果只剩下孤单的那一个元素
            single ^= num; 
        }
        return single;
    }
}

simplify:

class Solution {
    
    
    // 简化
    public int singleNumber(int[] nums) {
    
    
        for (int i = 1; i < nums.length; i++) {
    
    
            nums[0] ^= nums[i];
        }
        return nums[0];
    }
}