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.

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第一个数决定在第几个[0,(n-1)!],[(n-1)!,2(n-1)!]……这样的区间内，第一个确定之后确定第二个……以此类推

public class Solution {
public String getPermutation(int n, int k) {
String ret = "";
int s = 1;
for (int i = n-1;i>1;i--){
s*=i;
}
List <Integer> ku = new LinkedList <>();
for(int i = 1;i<=n;i++){
ku.add(i);
}
k--;
for(int i = 1;i<=n;i++){
int tem = k/s;
k = k%s;
if(n-i!=0)s = s/(n-i);
ret+=""+ku.get(tem);
ku.remove(tem);
}
return ret;
}
}