Single Number

https://oj.leetcode.com/problems/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?

public class Solution {

    public int singleNumber(int[] A) {



        int number = 0;

        HashMap<Integer, Integer> map = new HashMap();

        for (int i = 0; i < A.length; i++) {

            if (map.get(A[i]) == null || map.get(A[i]) == 0) {

                map.put(A[i], 1);

            } else if (map.get(A[i]) == 1) {

                map.put(A[i], 2);

            }

        }



        Iterator<Integer> it = map.keySet().iterator();

        while (it.hasNext()) {

            Integer keyString = it.next();

            if (map.get(keyString) == 1)

                return keyString;

        }

        return number;

    }

}

1. 最复杂的方法是，两次遍历，时间复杂度O(n^2)。

2. 上面的解法是，放入map，每次放入时候判断get是不是已经有了，有了就值为2。最后取出value为1的。时间复杂度为O(n)。

3. 还有一个很tricky的解法。用异或的方法。思路就是每位bit出现2次就清零，所以可以不断异或运算得出最终结果。

public class Solution {

    public int singleNumber(int[] A) {

        int result = 0;

        for(int i = 0; i < A.length; i++){

            result = result ^ A[i];

        }

        return result;

    }

}