问题描述：

The set [1,2,3,…,n] contains a total ofn! 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 k^th permutation sequence.

Note: Given n will be between 1 and 9 inclusive.

基本思想：

根据k和n！之间的关系，第k个序列应该在哪些位置提升几个数字。

代码：

 //consider if k > n! ？  
    string getPermutation(int n, int k) {   //C++
        //the first
        vector<char> base(n,'0'); 
        for(int i = 1; i<=n; i++)
            base[i-1] = i +'0';
        
        vector<int> multi(n);
        int temp = 1;
        multi[0] = 1;
        for(int i =1; i < n; i++){
            temp *= i;
            multi[i] = temp;
        }
        
        k--;
        for(int i = 0; i< n-1; i++){
            int temp = k/multi[n-i-1];
            
            int pos = i+temp;
            char c = base[pos];
            
            for(int j =pos-1 ; j>=i; j-- )
                base[j+1] = base[j];
                
            base[i] = c;
            k = k%multi[n-i-1];
        }
        
        //to String
        string result = "";
        for(int i = 0; i < n; i++)
            result += (base[i]);
        return result;
    }

[leetcode]Permutation Sequence

原文：http://blog.csdn.net/chenlei0630/article/details/42087015