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):

"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.

预先把1-9的阶乘都算出来，剩下就可以简单粗暴了

class Solution {
public:

string getPermutation(int n, int k) {
// Start typing your C/C++ solution below
// DO NOT write int main() function
vector<int> per;
if(n==0) return"";
if(n==1) return"1";
per.resize(9);
per[0]=1;
k--;
for(int i=1;i<9;i++){
per[i]=per[i-1]*(i+1);
}
k=k%per[n-1];
vector<int> vvv;
vvv.resize(9);
string ret="";
for(int i=0;i<9;i++)vvv[i]=i+1;
for(int i=0;i<n-1;i++){
int ind=n-2-i;
int d=k/per[ind];
k=k%per[ind];
for(int j=0;j<9;j++){
if(vvv[j]!=-1){
if(d!=0)d--;
else{
ret+='0'+vvv[j];
vvv[j]=-1;
break;
}
}
}
}
for(int j=0;j<9;j++){
if(vvv[j]!=-1){
ret+='0'+vvv[j];
vvv[j]=-1;
break;
}
}
return ret;
}
};

View Code