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.

Difficulty: Medium

这题两个解法，第一个用nextPermutation，暴力算法，结果超时了。

第二个，通过一个巧妙地算法，比如第一位数的每个数的可能性为（n-1）!，通过一个sign[9]来记录每个数有没有用过，再通过(k-1)/(n-1)!来确定每一位是哪个数

解法一没通过就不上了

解法二：

int jc(int a)
{
if(a==0)
return 1;
int ans = 1;
for(int i = 1;i<=a;i++)
ans = ans * i;
return ans;
}

string getPermutation(int n, int k) {
string s;
int sign[9] = {0};
int index, n1 = n;

for(int i = 0;i<n1;i++)
{
index = (k-1)/jc(n-1);
int j = 0;
for(j = 0;j<n1;j++)
{
if(sign[j]==0)
index--;
if(index<0)
break;
}
sign[j]++;
s.push_back(j+'0'+1);
k = k%jc(n-1);
if(k ==0)
k = jc(n-1);
n--;

}
return s;

}