leetcode之Single Number

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<span style="font-family: Arial, Helvetica, sans-serif; background-color: rgb(255, 255, 255);">本题网址：https://leetcode.com/problems/single-number/</span>

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?

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题目大概意思就是找出这个数组中只出现一次而不是两次的数。本题要求时间复杂度为线性时间，并且不以增加内存来实现。

于是，有一种比较好的思路就是采用集合，将遇到的数字加进集合，当再次遇到时从集合中删掉这个数字，循环结束后，剩下的那个数就是答案。

嗯，上代码：

public class Solution {
public int singleNumber(int[] nums) {
HashSet<Integer> set = new HashSet<>();
for (int i = 0; i < nums.length; i++) {
if (set.isEmpty())
set.add(nums[i]);
else if (set.contains(nums[i]))
set.remove(nums[i]);
else
set.add(nums[i]);
}
Iterator<Integer> iterator = set.iterator();
Integer result = 0;
while (iterator.hasNext()) {
result = iterator.next();
}
return result;
}
}