Permutation Sequence (LeetCode)

Question:

https://leetcode.com/problems/permutation-sequence/

 

N个字符的permutation个数是N!，所以对于N个字符的数列，如果最开始的字符是第kth大的 (starting from 1)，则它可能的取值范围是 ((k-1)*((N-1)!) + 1) 到 (k*((N-1)!)。

所以对于N数列，当前值在数列中的index值(k)可以如下计算可得：

k = number / ((N-1)!)

 

class Solution {
public:
    string getPermutation(int n, int k) {
        
        vector<int> table(n, 0);
        BuildTable(table, n);
        
        vector<char> restChars;
        
        for (int i = 1; i <= n; i++)
        {
            restChars.push_back(i+'0');
        }
        
        string result;
        
        k = k-1; // starting from 0
        
        for (int i = 0; i < n-1; i++)
        {
            // k must < table[i]
            int indexToRemove = k/table[i+1];
            
            result.push_back(restChars[indexToRemove]);
            restChars.erase(restChars.begin()+indexToRemove);
            
            k = k%table[i+1];
        }
        
        result.push_back(restChars[0]);
        
        return result;
    }
    
    void BuildTable(vector<int>& table, int n)
    {
        table[n-1] = 1;
        
        int i = 2;
        int index = n-2;
        while (i <= n)
        {
            table[index] = table[index+1] * i;
            
            i++;
            index--;
        }
    }
};