标签：htm tput dfs input ica 思路 i++ add dex
Given a collection of integers that might contain duplicates, nums, return all possible subsets (the power set).
Note: The solution set must not contain duplicate subsets.
Input: [1,2,2] Output: [ [2], [1], [1,2,2], [2,2], [1,2], [] ]
题意：
是的[leetcode]78. Subsets数组子集 follow up
这题强调给定数组可以有重复元素
思路：
需要先对给定数组排序，以便去重
代码：
1 class Solution { 2 public List<List<Integer>> subsetsWithDup(int[] nums) { 3 List<List<Integer>> result = new ArrayList<>(); 4 if(nums == null || nums.length ==0 )return result; 5 Arrays.sort(nums); 6 List<Integer> path = new ArrayList<>(); 7 dfs(0, nums, path, result); 8 return result; 9 } 10 11 private void dfs(int index, int[] nums, List<Integer> path, List<List<Integer>> result){ 12 result.add(new ArrayList<>(path)); 13 for(int i = index; i < nums.length; i++){ 14 if( i != index && nums[i] == nums[i-1] ) continue; 15 path.add(nums[i]); 16 dfs(i + 1, nums, path, result); 17 path.remove(path.size() - 1); 18 } 19 } 20 }
[leetcode]90. Subsets II数组子集(有重)
标签：htm tput dfs input ica 思路 i++ add dex
原文地址：https://www.cnblogs.com/liuliu5151/p/9191503.html
