Leetcode Note: Permutation Sequence

Leetcode Note: Permutation Sequence

I. Description

Assume that {1, 2, 3, 4 ,..., N}, arrange the elements in it. There are n in total! And sort them from small to large. What is the form of the k combination? <喎?http: www.bkjia.com kf ware vc " target="_blank" class="keylink"> Memory + memory/O94qGjvt/M5cq1z9a/ybLO1dXM4sS/memory + 1xMrHt723qLb + Memory + memory/memory + Memory + 6/memory + Memory = = "brush: java; "> X=an*(n-1)!+an-1*(n-2)!+...+ai*(i-1)!+...+a2*1!+a1*0!

For example:

Question1324Yes{1,2,3,4}The number of combinations in the permutation:

First1, Less1No, yes0,0*3!The second digit is3, Less3The number1And2,1It already exists first, so there is only one number2,1*2!. The third digit is2Less2The number is1,1In the first place, so there are0Number,0*1!, So1324Small arrays have0*3!+1*2!+0*1!=2,1324Yes3.

The above is the Cantor encoding process, that isMap a fully ordered ing 1324 to a natural number 3The question is a known natural number.kThe corresponding full arrangement is a decoding process relative to the above steps. The following is a specific example:

How to find out{1,2,3,4,5}The16?
1. Use16-1, Get15;
2. Use15Remove4!, Get0, Yu15;
3. Use15Remove3!, Get2, Yu3;
4. Use3Remove2!, Get1, Yu1;
5. Use1Remove 1! , Get1, Yu0;
6. Yes0A smaller number than it is1So the first priority is1;
7. Yes2A smaller number than it is3,1It already exists before, so the second is4;
8. Yes1A smaller number than it is2,1So the third digit is3;
9. Yes1A smaller number than it is2But 1, 3, and 4 have all appeared, so the fourth digit is5;
10. Based on the above inference, the last number can only be2;

Therefore{1,4,3,5,2}.

Based on the above ideas, you can start designing algorithms.

Iii. Sample Code

#include 
  
   #include 
   
    #include 
    
     using namespace std;class Solution{public:    string PermutationSequence(int n, int k)    {        int total = CombinedNumber(n - 1);        if (k > total)        {            cout << The k is larger then the total number of permutation sequence: << total << endl;            return Null!;        }        string a(n, '0');        for (int i = 0; i < n; ++i)            a[i] += i + 1;   // sorted        // Cantor expansion        string s = a, result;        k--; // (k - 1) values are less than the target value         for (int i = n - 1; i > 0; --i)        {            auto ptr = next(s.begin(), k / total);            result.push_back(*ptr);            s.erase(ptr);  // delete the already used number            k %= total;    // update the dividend            total /= i;    // update the divider        }        result.push_back(s[0]);  // The last bit        return result;    }private:    int CombinedNumber(int n)    {        int num = 1;        for (int i = 1; i < n + 1; ++i)            num *= i;        return num;    }};
    
   
  

The following is a simple test code:

#include PermutationSequence.hint main(){    Solution s;    int n = 6, k = 150;    string result = s.PermutationSequence(n, k);    std::cout << n =  << n <<  and the  << k << th sequence is:  << result << std::endl;    getchar();    return 0;}

A correct test result,n = 6,k = 16:

WhenkWhen the value of exceeds the number of possible combinations: