标签：nts int algorithm single i++ rect get and elements
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.
Example:
Input:[1,2,1,3,2,5]Output:[3,5]
Note:
[5, 3] is also correct.class Solution { public int[] singleNumber(int[] nums) { List<Integer> list = singleNumber1(nums); int[] res = new int[list.size()]; for(int i = 0; i < list.size(); i++){ res[i] = list.get(i); } return res; } public List<Integer> singleNumber1(int[] nums) { List<Integer> list = new ArrayList(); Map<Integer, Integer> map = new HashMap(); for(int i : nums){ int value = map.getOrDefault(i, 0); map.put(i, value + 1); } for(Map.Entry<Integer, Integer> entry : map.entrySet()){ if(entry.getValue() == 1) list.add(entry.getKey()); } return list; } }
调用1中map方法，屡试不爽
class Solution { public int[] singleNumber(int[] nums) { if(nums == null || nums.length == 0) return new int[0]; Map<Integer, Integer> map = new HashMap<>(); for(int i = 0; i < nums.length; i++) { if(!map.containsKey(nums[i])) { map.put(nums[i], 1); }else{ map.remove(nums[i]); } } int[] output = new int[2]; int n = 0; for(int i : map.keySet()) { output[n++] = i; } return output; } }
或者稍微优化一下。
标签：nts int algorithm single i++ rect get and elements
原文地址：https://www.cnblogs.com/wentiliangkaihua/p/11702705.html
