LeetCode 47. Permutations II(生成不同的组合-dfs)
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Given a collection of numbers that might contain duplicates, return all possible unique permutations.
For example,[1,1,2]
have the following unique permutations:
[ [1,1,2], [1,2,1], [2,1,1]]
用的是dfs来完成
public class Solution { public List<List<Integer>> permuteUnique(int[] nums) { if(nums==null||nums.length==0) return null;Arrays.sort(nums); //把重复数字放在一起 int len = nums.length; int[] isUsed = new int[len]; List<List<Integer>> result = new ArrayList<List<Integer>>(); ArrrayList<Integer> store = new ArrayList<Integer>(); dfs(result,isUsed,store,nums); return result; } public void dfs(List<List<Integer>> result, int[] isUsed, List<Integer> store, int[] nums) { if(store.size()==nums.length){ result.add(new ArrayList<Integer>(store)); return; } for(int i=0;i<nums.length;i++){ if(isUsed[i]==1) continue; if(i>0&&nums[i]==nums[i-1]&&isUsed[i-1]==0) continue; //重复的当前面使用过才可以使用,前面未使用过直接使用会导致解重复 store.add(nums[i]); isUsed[i]=1; dfs(result,isUsed,store,nums); isUsed[i]=0; store.remove(store.size()-1); }}}
result.add(new ArrayList<Integer>(store));这个有问题等待补充!
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