LeetCode——Single Number III

Description：

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:

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

在线性时间复杂度下，找到两个只出现一次的元素。结果没有先后顺序。

既然是只出现一次且不考虑顺序，那么就可以机智的使用set的无序不可重复的特性。

public class Solution {
    public int[] singleNumber(int[] nums) {
        Set<Integer> set = new HashSet<Integer>();
        
        for(int i : nums) {
            if(!set.add(i)) {
                set.remove(i);
            }
        }
        int[] res = new int[2];
        int i = 0;
        for(int e : set) {
            res[i++] = e;
        }
        
       return res;
    }
}

LeetCode——Single Number III

原文：http://www.cnblogs.com/wxisme/p/4842896.html