LeetCode——Permutation Sequence

最新推荐文章于 2019-05-12 11:52:31 发布

yjfanling

最新推荐文章于 2019-05-12 11:52:31 发布

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分类专栏：算法文章标签： java leetcode

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本文介绍了一种高效算法，用于解决寻找从1到n的所有唯一排列中第k个排列的问题。通过递归分解和数学组合，算法在O(n)时间内生成结果，适用于n在1到9之间的场景。

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The set [1,2,3,…,n] contains a total of n! unique permutations.
By listing and labeling all of the permutations in order, we get the following sequence for n = 3:
“123”
“132”
“213”
“231”
“312”
“321”
Given n and k, return the kth permutation sequence.
Note:
Given n will be between 1 and 9 inclusive.
Given k will be between 1 and n! inclusive.

Example 1:
Input: n = 3, k = 3
Output: “213”
Example 2:
Input: n = 4, k = 9
Output: “2314”

解法

这道题就是找规律，具体怎么找可以自己画一组数跟着代码过一遍流程就懂了，语言能力不行表述不清，就不误导大家了，还是自己想更能想明白

public String getPermutation(int n, int k) {
        List<Integer> list=new ArrayList<Integer>();
        for(int i=1;i<=n;i++)
        	list.add(i);
        int an=(k-1)/f(n-1);
        int kn=(k-1)%f(n-1);
        String res=String.valueOf(list.get(an));
        list.remove(an);
        for(int i=2;i<=n;i++)
        {
        	an=kn/f(n-i);
        	kn=(k-1)%f(n-i);
        	res=res+String.valueOf(list.get(an));
        	list.remove(an);
        }
        return res;
	}
	private int f(int i) {
		// TODO Auto-generated method stub
		int res=1;
		for(int j=2;j<=i;j++)
			res*=j;
		return res;
	}
}

Runtime: 7 ms, faster than 80.89% of Java online submissions for Permutation Sequence.
Memory Usage: 37.6 MB, less than 33.33% of Java online submissions for Permutation Sequence.