leetcode60 - Permutation Sequence - hard

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 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.
• Given k will be between 1 and n! inclusive.

Example 1:

Input: n = 3, k = 3
Output: "213"

Example 2:

Input: n = 4, k = 9
Output: "2314"

 

看了最优解做的。大致是把所有的permutations break town来思考。比如说如果我们确定了最高位，那可以分成三组（分别是首位为1or2or3），每组2个。接下来每组中如果确定第二位，可以分成两组，每组1个，以此类推，这样就每次缩小了寻找范围。

首先，确定第i （0~n-1）位后，当前这个大group就可以被拆分成每组有npg=factorial(n-1-i)个数的小groups。那kth就落在第k/npg个组，接下来我们就在这个组里找第k%npg个数。确认一个数字的位置后，要把标成不可再用。

 

细节：

int to char: '0'+c;

加进string里 res[i]=x不行，用res+=x

 

实现：Time O(n^2) Space O(n)

class Solution {
public:
    string getPermutation(int n, int k) {
        
        vector<int> factorial(n);
        factorial[0] = 1;
        for (int i=1; i<n; i++){
            factorial[i] = factorial[i-1]*i;
        }
        
        vector<int> available;
        for (int i=1; i<=n; i++){
            available.push_back(i);
        }
        
        k-=1;
        string res = "";
        for (int i=0; i<n; i++){
            int numPerGroup = factorial[n-1-i];
            int group = k/numPerGroup;
            int cur = available[group];
            res += '0'+ cur;
            available.erase(find(available.begin(), available.end(), cur));
            k %= numPerGroup;
        }
        
        return res;
        
    }
};