[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):

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.

 

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  Backtracking Math

 

 

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    这道题可以直接穷举然后数第k 个，或者 迭代k次得到返回值，或者直接计算出结果。我用是直接计算结果，很多细节上需要处理。

 

#include <iostream>

#include <string>

using namespace std;



class Solution {

public:

    string getPermutation(int n, int k) {

        if(n <2)    return "1";

        int tab[10]={1};

        for(int i =1;i<=9;i++)

            tab[i] = tab[i-1]*i;

        int remain[9]={0};

        for(int i=0;i<9;i++)

            remain[i]=i+1;

        int cnt = n;

        string ret = "";

        while(cnt>0){

            cnt--;

            int num = (k-1)/tab[cnt];

//            cout<<"num="<<num<<endl;

            int idx = 0;

            while(remain[idx]==0)   idx++;

            for(int i=0;i<num;i++)

                if(remain[++idx]==0)  i--;

//            cout<<"idx="<<idx<<endl;

            ret += '0'+remain[idx];

            remain[idx]=0;

            k = k - num*tab[cnt];

//            cout<<endl;

        }

        return ret;

    }

};



int main()

{

    Solution sol;

    cout<<sol.getPermutation(9,213)<<endl;

    return 0;

}