leetcode之Single Number
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<span style="font-family: Arial, Helvetica, sans-serif; background-color: rgb(255, 255, 255);">LeetCode官网网址:https://leetcode.com</span>
<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; }}
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