[leetcode] Permutation Sequence

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.

 

Solution:

计算 1~n 数字的第k个排列

思路1：从小到大生成同时计数，直到第k个。 肯定超时试都不用试。。

思路2：直接根据规律计算第k个排列。

分析：1~n个数共有n!个排列，1开头的有(n-1)!个，2开头的(n-1)!个，...n开头的有(n-1)!个。因此用k/(n-1)!就确定了第一位数字，然后依次类推（用过的数字要去除），继续在(n-1)!个数中找第k%(n-1)!个数。

 

public class Solution {
    public String getPermutation(int n, int k) {
        String s_n = generateString(n);
        int total = factorial(n);
        if(n==1 && k==1)
            return "1";

        char[] result = new char[n];

        for (int i = 0; i < n; ++i) {
            total /= (n - i);
            int index = (k - 1) / total;
            result[i] = s_n.charAt(index);
            
            s_n=s_n.replace(result[i]+"", "");
            k-=index*total;
        }

        return new String(result);
    }

    private String generateString(int n) {
        // TODO Auto-generated method stub
        String sb = "";
        for (int i = 1; i <= n; ++i) {
            sb += i + "";
        }
        return sb;
    }

    private int factorial(int n) {
        // TODO Auto-generated method stub
        int total = 1;
        for (int i = 1; i <= n; i++)
            total *= i;
        return total;
    }
}