[Leetcode] permutation sequence

The set [1, 2, 3 ,...,N] Contains a totalN! Unique permutations.

By listing and labeling all of the permutations in order,
We get the following sequence (ie,N= 3 ):

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

GivenNAndK, ReturnKTh permutation sequence.

Https://oj.leetcode.com/problems/permutation-sequence/

Computing 1 ~ K of N numbers

Idea 1: Count from small to large until the K. It must be time-out ..

Idea 2: Calculate the k-th order based on the rule.

Analysis: 1 ~ N total N! (N-1 )! , Starting with 2 (n-1 )! (N-1) at the beginning of... n )! . Therefore, K/(n-1) is used )! The first number is determined, and the number is removed in turn (n-1 )! K % (n-1) in the number )! Number.

Idea 2 code:

 

public class Solution {    public String getPermutation(int n, int k) {        int[] num = new int[n];        int permSum = 1;        for (int i = 0; i < n; i++) {            num[i] = i + 1;            permSum *= (i + 1);        }        StringBuilder sb = new StringBuilder();        k--;//change to base 0        for (int i = 0; i < n; i++) {            permSum = permSum / (n - i);            int selected = k / permSum;            sb.append(num[selected]);            for (int j = selected; j < n - i - 1; j++)                num[j] = num[j + 1];            k = k % permSum;        }        return sb.toString();    }    public static void main(String[] args) {        System.out.println(new Solution().getPermutation(4, 10));    }}

 

Refer:

Http://blog.csdn.net/havenoidea/article/details/12837441

Http://www.cnblogs.com/TenosDoIt/p/3721918.html