136. Single Number

题目

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?

分析

数组中所有数都出现2次，除了某个数只出现1次，求出这个数

解答

解法1：（我）HashMap，使用了额外的内存（15ms ×）

key为元素值，value为1，重复的元素删掉，最后只剩下一个元素

public class Solution {
    public int singleNumber(int[] nums) {
        HashMap hashMap = new HashMap<Integer, Integer>(((int)(nums.length / 0.75) < 16) ? 16 : (int)(nums.length / 0.75));
        int result = 0;
        for (int num : nums){
            if(hashMap.containsKey(num)){
                hashMap.remove(num);
            }
            else{
                hashMap.put(num,1);
            }
        }
        Set set = hashMap.keySet();
        Iterator iterator = set.iterator();
        while(iterator.hasNext()){
            result = (Integer)iterator.next();
        }
        return result;
        
    }
}

解法2：二进制异或运算（1ms√）

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

所以，如果N是single number，

N1 ^ N1 ^ N2 ^ N2 ^{..............} Nx ^ Nx ^ N

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

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

= N

public class Solution {
    public int singleNumber(int[] nums) {
        int result = 0;
        for (int num : nums){
            result ^= num;
        }
        return result;
        
    }
}