【LeetCode】060. Permutation Sequence

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.

题解：

Solution 1

class Solution {
public:
    string getPermutation(int n, int k) {
        string s;
        for(int i = ; i < n; ++i){
            s += (i + ) + '';
        }
        for(int i = ; i < k - ; ++i){
            next_permutation(s);
        }
        return s;
    }
    void next_permutation(string &str){
        int n = str.size();
        for(int i = n - ; i >= ; --i){
            if(str[i] >= str[i + ]) continue;
            int j = n - ;
            for(; j > i; --j) {
                if(str[j] > str[i]) break;
            }
            swap(str[i], str[j]);
            reverse(str.begin() + i + , str.end());
            return;
        }
        reverse(str.begin(), str.end());
    }
};

Solution 2

class Solution {
public:
    string getPermutation(int n, int k) {
        string res;
        if(n <=  || k <= ){
            return res;
        }
        string num = "";
        vector<int> f(n, );
        for(int i = ; i < n; ++i){
            f[i] = f[i - ] * i;
        }
        --k;
        for(int i = n; i > ; --i){
            int j = k / f[i - ];
            k %= f[i - ];
            res.push_back(num[j]);
            num.erase(j, );
        }
        return res;
    }
};

康托编码

Solution 3

class Solution {
public:
    string getPermutation(int n, int k) {
        string s = "", str;
        int factorial = ;
        for(int i = ; i < n; ++i){
            factorial *= i;
        }
        --k;
        for(int i = n; i > ; --i){
            int index = k / factorial;
            str += s[index];
            s.erase(index, );
            k %= factorial;
            factorial /= i -  ? i -  : ;
        }
        return str;
    }
};