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# 題目敘述

Given an integer array nums , in which exactly two elements appear only once and all the other elements appear exactly twice. Find the two elements that appear only once. You can return the answer in any order.

You must write an algorithm that runs in linear runtime complexity and uses only constant extra space.

# Example 1

Input: nums = [1,2,1,3,2,5]
Output: [3,5]
Explanation: [5, 3] is also a valid answer.

# Example 2

Input: nums = [-1,0]
Output: [-1,0]

# Example 3

Input: nums = [0,1]
Output: [1,0]

# 解題思路

# Solution

Solution.java
import java.util.HashMap;
import java.util.Map;
class Solution {
    public int[] singleNumber(int[] nums) {
        Map<Integer, Integer> map = new HashMap<>();
        for (int num : nums) {
            map.put(num, map.getOrDefault(num, 0) + 1);
        }
        int[] result = new int[2];
        int index = 0;
        for (Map.Entry<Integer, Integer> entry : map.entrySet()) {
            if (entry.getValue() == 1) {
                result[index++] = entry.getKey();
            }
        }
        return result;
    }
}
Solution.java
class Solution {
    public int[] singleNumber(int[] nums) {
        int xor = 0;
        for (int num : nums) xor ^= num;
        int mask = 1;
        while ((xor & mask) == 0) mask <<= 1;
        int a = 0;
        for (int num : nums) {
            if ((num & mask) == 0) a ^= num;
        }
        return new int[] {a, xor ^ a};
    }
}




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