136. Single Number
Given an array of integers, every element appears twice except for one. Find that single one.
Bit: a ^ a = 0;
public class Solution { public int singleNumber(int[] nums) { int res = 0; for(int n : nums){ res ^= n; } return res; } }
137. Single Number II
public class Solution { public int singleNumber(int[] nums) { HashMap<Integer, Integer> map = new HashMap<>(); for(int i = 0 ; i < nums.length ; i++){ map.put(nums[i], map.getOrDefault(nums[i], 0) + 1); } for(Map.Entry<Integer, Integer> entry : map.entrySet()){ if(entry.getValue() == 1) return entry.getKey(); } return 0; } }
260. Single Number III
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].
public class Solution { public int[] singleNumber(int[] nums) { Set<Integer> set = new HashSet<>(); for(int n: nums){ if(!set.add(n)){ set.remove(n); } } Iterator<Integer> it = set.iterator(); int[] res = new int[2]; int start = 0; while(it.hasNext() && start < res.length){ res[start] = it.next(); start++; } return res; } }
二刷可以研究bit做法
原文：http://www.cnblogs.com/joannacode/p/6121399.html