260.Single Number III
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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?
public class Solution { /* * 利用HashSet的唯一性 */ public int[] singleNumber(int[] nums) { HashSet<Integer> hash = new HashSet<>(); for(int n: nums){ if(!hash.add(n)){ hash.remove(n); } } int i = 0; int[] array = new int[hash.size()]; for(int b: hash){ array[i++] = b; } return array; }}
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- 260.Single Number III
- 260. Single Number III
- 260. Single Number III
- 260. Single Number III
- 260. Single Number III
- 260. Single Number III
- 260. Single Number III
- 260. Single Number III
- 260. Single Number III
- 260. Single Number III
- 260. Single Number III
- 260. Single Number III
- 260. Single Number III
- 260. Single Number III
- 260. Single Number III
- 260. Single Number III
- 260. Single Number III
- 260. Single Number III
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