[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.

最开始是暴力算，但是9！实在是太大了，显然超时，后来网上看到大牛们的算法，从前后向后找，已知对于每一次第n位的数字，都有(n-1)!个排列，所以可以根据这个规律直接找到答案。

class Solution {  
public:  
    string getPermutation(int n, int k) {
        string str;
        vector<int> factorial(n + 1, 1);
        for (int i = 1; i <= n; ++i) {
            str += i + '0';
            factorial[i] = factorial[i-1] * i;
        }
        string perm;
        --k;    // convert to 0-based index
        for (int i = n - 1; i >= 0; --i) {
            int quotient = k / factorial[i];
            perm += str[quotient];
            str.erase(quotient, 1);
            k %= factorial[i];
        }
        return perm;
    }
}; 

超时的算法：

class Solution {
public:
    string getPermutation(int n, int k) {
        string str;
        for (int i = 1; i <= n; ++i) {
            str.push_back('0' + i);
        } 
        if (n < 2) return str;
        
        int a, b;
        while (k--) {
            for (a = n - 2; a >= 0; --a) 
                if (str[a] < str[a+1]) 
                    break;
            
            for (b = n - 1; b > a; --b) 
                if (str[b] > str[a])
                    break;
            
            swap(str[a], str[b]);
            reverse(str.begin() + a + 1, str.end());
        }
        return str;
    }
};