Leetcode Single Number

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?

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解题思路：

1. 瞬间想到用HashSet, 但是时间复杂度也是可想而知。

2. 参考答案发现巧妙的方法，使用异或(XOR)^  速度超快！！！

we use bitwise XOR to solve this problem :

first , we have to know the bitwise XOR in java

1. 0 ^ N = N
2. N ^ N = 0

So..... if N is the single number

N1 ^ N1 ^ N2 ^ N2 ^..............^ Nx ^ Nx ^ N

= (N1^N1) ^ (N2^N2) ^..............^ (Nx^Nx) ^ N

= 0 ^ 0 ^ ..........^ 0 ^ N

= N

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Java code:

方法一： HashSet(不推荐，速度慢)

public int singleNumber(int[] nums) {
        Set<Integer> set = new HashSet<Integer>();
        for(int i = 0; i< nums.length; i++){
            if(set.contains(nums[i])){
                set.remove(nums[i]);
            }else{
                set.add(nums[i]);
            }
        }
        Iterator it = set.iterator();
        int result = (int)it.next();
        return result;
    }

方法二：

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

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Reference:

1. https://leetcode.com/discuss/53327/easy-java-solution-tell-you-why-using-bitwise-xor