[LeetCode][Java] Single Number
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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?
题目大意:
给定一个整数数组,数组中每个元素都显示两次,只有一个元素只显示一次。找出这个只显示一次的元素。
注意:
要求线性时间复杂度。
分析:
最易想到的做法是,把每个元素一次存储到HashMap中,出现重复的就删除,最后剩余的那个元素就是要找的只出现一次的元素。
代码如下:
public class Solution { public int singleNumber(int[] nums) {Map<Integer,Integer> input=new HashMap<Integer,Integer>(); input.put(nums[0], 0); int finalnum = 0; int label=0;for(int i=1;i<nums.length;i++) { if(input.containsKey(nums[i])) input.remove(nums[i]); else input.put(nums[i],i); }for(int i=0;i<nums.length;i++) { if(input.containsKey(nums[i])) { label=input.get(nums[i]); finalnum=nums[label]; break; } } return finalnum; }}
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