Leetcode 136. Single Number

Description:

Given an array of integers, every element appears twice except for one. Find that single one.

Note: Your algorithm should have a linear runtime complexity. Could you implement it without using extra memory?

Thinking:

Knowing that A XOR A = 0 and the XOR operator is commutative, the solution will be very straightforward.

Complexity:

The time complexity is O(n).

Step:

1. check whether input is valid.
2. Do a for loop and use XOR.

Code:

public class Solution {
    public int singleNumber(int[] nums) {
        if(nums == null || nums.length == 0) return 0;
        int result = 0;
        for(int i = 0; i < nums.length; i++){
            result^= nums[i];
        }
        return result;
    }
}

Conclusion:

we have to know the bitwise XOR in java 0 ^ N = N,N ^ N = 0.