題目：
Given an array of numbers 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.

For example:

Given nums = [1, 2, 1, 3, 2, 5], return [3, 5].

Note:
The order of the result is not important. So in the above example, [5, 3] is also correct.
Your algorithm should run in linear runtime complexity. Could you implement it using only constant space complexity?

翻譯：
給定一個數組，里面有2個元素，只出現一次，其他元素都是出現2次。求這2個出現一次的元素。
說明：返回結果的數組順序不重要。你的程序需要運行在線性復雜度下。你可以實現它只用常數級的空間復雜度嗎？

代碼：

public class Solution {public int[] singleNumber(int[] nums) {int[] result=new int[2];Map<Integer,Integer> map=new HashMap<>();for(int i=0;i<nums.length;i++){Integer value=map.get(nums[i]);if(value==null){map.put(nums[i],1);}else{map.put(nums[i],1+value);}}int resultIndex=0;for(Integer key:map.keySet()){if(map.get(key)==1){result[resultIndex]=key;resultIndex++;}}return result;} }

總結

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