【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 = 0; i < n; ++i){
            s += (i + 1) + '0';
        }
        for(int i = 0; i < k - 1; ++i){
            next_permutation(s);
        }
        return s;
    }
    void next_permutation(string &str){
        int n = str.size();
        for(int i = n - 2; i >= 0; --i){
            if(str[i] >= str[i + 1]) continue;
            int j = n - 1;
            for(; j > i; --j) {
                if(str[j] > str[i]) break;       
            }
            swap(str[i], str[j]);
            reverse(str.begin() + i + 1, str.end());
            return;
        }
        reverse(str.begin(), str.end());
    }
};

Solution 2 

class Solution {
public:
    string getPermutation(int n, int k) {
        string res;
        if(n <= 0 || k <= 0){
            return res;
        }
        string num = "123456789";
        vector<int> f(n, 1);
        for(int i = 1; i < n; ++i){
            f[i] = f[i - 1] * i;
        }
        --k;
        for(int i = n; i > 0; --i){
            int j = k / f[i - 1];
            k %= f[i - 1];
            res.push_back(num[j]);
            num.erase(j, 1);
        }
        return res;
    }
};

康托编码

Solution 3

class Solution {
public:
    string getPermutation(int n, int k) {
        string s = "123456789", str;
        int factorial = 1;
        for(int i = 1; i < n; ++i){
            factorial *= i;
        }
        --k;
        for(int i = n; i > 0; --i){
            int index = k / factorial;
            str += s[index];
            s.erase(index, 1);
            k %= factorial;
            factorial /= i - 1 ? i - 1 : 1;
        }
        return str;
    }
};