LeetCode OJ:Permutation Sequence

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.

深度暴力果然超时

class Solution {
public:
string dfs(int num[],int m,int n,int &cnt,int k){
if(m>n){
if(++cnt==k){
string str;
for(int i=1;i<=n;i++)
str+=(char)(num[i]+'0');
return str;
}
return "";
}
for(int i=m;i<=n;i++){
swap(num[i],num[m]);
string t=dfs(num,m+1,n,cnt,k);
if(t!="")return t;
swap(num[i],num[m]);
}
return "";
}
string getPermutation(int n, int k) {
int num[10];
for(int i=1;i<=9;i++)num[i]=i;
int cnt=0;
return dfs(num,1,n,cnt,k);
}
};

看来得找规律了，

第一位每个数字开头的序列都有（n-1）！个序列；

第二位每一个数开头都有（n-2）！个序列；依次类推

康托编码：

假设n个不重复元素，第k个排列是a1,a2,a3,a4,a5...an

a1 = k/(n-1)!

k2 = k%(n-1)!

a2 = k2/(n-2)!

........

........

k(n-1) = k(n-2)%2!

a(n-1) = k(n-1)/1!

an = 0

class Solution {
public:
string getPermutation(int n, int k) {
string result;
string s(n,'0');
int i,d(1),q;
for(i=1;i<=n;++i){
s[i-1] += i;
d *= i;
}
--k;//k--因为康托编码从0开始
for(i=n;i>=1;--i){
d = d/i;
q = k/d;
k %=d;
result.push_back(s[q]);
s.erase(s.begin()+q);
}
return result;
}
};