leetcode 60. 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):

"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.

按字典序全排列组合，求第k个组合是什么

思路是，对每一个数求它在这个排列中的顺序是怎么的，从而得到这一个位置的数字

public class Solution {
public String getPermutation(int n, int k) {
k--;
StringBuilder sb = new StringBuilder();
int factorial = 1;
for(int i = 2; i < n; i++)
factorial *= i;
List<Integer> numList = new ArrayList<Integer>();
for(int i = 1; i<= n; i++){
numList.add(i);
}
int times = n-1;
while(times>=0){
int index = k/factorial;
sb.append(numList.get(index));
numList.remove(index);
k = k%factorial;
if(times!=0)
factorial = factorial/times;
times--;
}
return sb.toString();
}
}