【Permutation Sequence】cpp

题目：

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):

1. "123"
2. "132"
3. "213"
4. "231"
5. "312"
6. "321"

 

Given n and k, return the kth permutation sequence.

Note: Given n will be between 1 and 9 inclusive.

代码：

class Solution {
public:
    string getPermutation(int n, int k)
    {
        std::stringstream result;
        // if the digit has been used
        int flag[10];
        for(int i = 1; i<10; ++i) flag[i]=1;
        // calculate each digit
        int factorical,quotient,remaind=k-1;
        for (int i = 1; i<=n; ++i)
        {
             factorical = Solution::getFactorical(n-i);
            quotient = remaind/factorical;
            int tmp = 0;
            for(int j=1; j<=9; ++j){
                tmp = tmp + flag[j];
                if ( tmp==(quotient+1) ){
                    quotient = j;
                    break;
                }
            }
            result << quotient;
            flag[quotient] = 0;
            remaind = remaind%factorical; // update remaind
        }
        return result.str();
    }
    static int getFactorical(int n){
        int result = 1;
        for (int i = 1; i <= n; ++i){
            result = result * i;
        }
        return result;
    }
};

Tips:

1. 主要思路是康托编码，具体什么是康托编码（http://blog.sina.com.cn/s/blog_4bf7b6580100l2zs.html）

2. 把计算阶乘的代码单独列出来

3. 对于c++如何将int转为字符不太懂，因此用了sstream的这个方法，后面遇到其他的方法再改进一下。

======================================================

第二次过这道题，对康托编码的算法有些生疏了，参考了blog才回忆起来。然后扫了一些细节，形成了自己的代码。

class Solution {
public:
        string getPermutation(int n, int k)
        {
            vector<char> ret;
            vector<bool> visit(9,false);
            k = k-1;
            for ( int i=n-1; i>=0; --i)
            {
                int factor = Solution::factorial(i);
                int num_less = k / factor;
                // find num less
                int j=0;
                int count = 0;
                for ( ; j<9; ++j )
                {
                    if ( !visit[j] )
                    {
                        if ( count==num_less ) break;
                        count++;
                    }
                }
                visit[j] = true;
                ret.push_back(j+'1');
                k = k % factor; 
            }
            return string(ret.begin(),ret.end());
        }
        static int factorial(int n)
        {
            if ( n==0 || n==1 ) return 1;
            int ret = 1;
            for ( int i = 1; i<=n; ++i) ret = ret * i;
            return ret;
        }
};