Permutation Sequence

Permutation Sequence

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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 (ie, for n = 3):

1. "123"
2. "132"
3. "213"
4. "231"
5. "312"
6. "321"

Given n and k, return the kth permutation sequence.

Note: Given n will be between 1 and 9 inclusive.

Have you been asked this question in an interview? 

假设有n个元素，第K个permutation是
a1, a2, a3, .....   ..., an
那么a1是哪一个数字呢？

那么这里，我们把a1去掉，那么剩下的permutation为
a2, a3, .... .... an, 共计n-1个元素。 n-1个元素共有(n-1)!组排列，那么这里就可以知道

设变量K1 = K
a1 = K1 / (n-1)!// 第一位的选择下标

同理，a2的值可以推导为

K2 = K1 % (n-1)!
a2 = K2 / (n-2)!

。。。。。

K(n-1) = K(n-2) /2!
a(n-1) = K(n-1) / 1!

an = K(n-1)

public class Solution {
    public String getPermutation(int n, int k) {
        int data[]=new int[10];
        boolean visited[]=new boolean[10];
        data[0]=data[1]=1;
        ArrayList<Integer> list=new ArrayList<Integer>();
        for(int i=1;i<=n;i++)
        {
            data[i]=data[i-1]*(i);
            list.add(i);
        }
        String result="";
        k--;
        for(int i=n-1;i>=0;i--)
        {
            int cur=k/data[i];
            int j=1;
            for(;j<9;j++)　　　　　//从数组中 针对未访问过的元素visited[]=false  找第cur个,找到的为j
            {
                if(visited[j]==false)
                    cur--;
                if(cur<0)
                    break;
            }
            visited[j]=true;
            result+=(j);
            k=k%data[i];
        }
        return result;
    }
    
}

解二

public class Solution {
    public String getPermutation(int n, int k) {
        int data[]=new int[n+1];
        boolean visited[]=new boolean[n+1];
        data[0]=data[1]=1;
        ArrayList<Integer> list=new ArrayList<Integer>();
        for(int i=1;i<=n;i++)
        {
            data[i]=data[i-1]*(i);
            list.add(i);
        }
        String result="";
        k--;
        for(int i=n-1;i>=0;i--)
        {
            int cur=k/data[i];
            result+=list.remove(cur);        //第cur个未正解，从list中删除
            k=k%data[i];
        }
        return result;
    }
}

基本思路就是