47 Permutations II

最新推荐文章于 2024-07-14 23:58:08 发布

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最新推荐文章于 2024-07-14 23:58:08 发布

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本文介绍了一种使用回溯法生成包含重复元素数组的所有唯一排列的方法，并通过排序和标记已使用元素的方式避免重复解的产生。

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分析摘自：

https://leetcode.com/discuss/18482/share-my-recursive-solution

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], and [2,1,1].

回溯法生成全排列+hash判重，可过，不优。

相同的元素为【源】调用生成重复集合，没有顺序？就规定顺序。

排序后将相同的元素和前面的元素【捆绑】以免重复调用。

The idea to resolve duplicate is to ensure that for elements with the same value, we make sure that they are picked up in the ascending order of index. To implement this, every time we try to pick up some value, we just check if the previous element has the same value and is visited or not. If so, we just return!

List<List<Integer>> list=new ArrayList<>();
	ArrayList<Integer> arraylist=new ArrayList<>();
	boolean[] onstack;
	
	public List<List<Integer>> permuteUnique(int[] nums) 
	{
		Arrays.sort(nums);
		onstack=new boolean[nums.length];
		dfs(nums.length, 0, nums);
	
		return list;
	}
	
	
	
	public void dfs(int n,int k,int[] nums)
	{
		if(k==n)
		{
			list.add(new ArrayList<>(arraylist));
			return ;
		}
		
		for(int i=0;i<nums.length;i++)
		{
			if(!onstack[i])
			{
				if(i>0 &&nums[i-1]==nums[i] && !onstack[i-1])
					continue;
				
				arraylist.add(nums[i]);
				onstack[i]=true;
				dfs(nums.length, k+1 , nums);
				onstack[i]=false;
				arraylist.remove(arraylist.size()-1);
			}
		}
	}