leetcode 60. Permutation Sequence(康托展开)

描述：

　　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. 1. 　　"123"
   2. 　　"132"
   3. 　　"213"
   4. 　　"231"
   5. 　　"312"
   6. 　　"321"

　　Given n and k, return the kth permutation sequence.

思路：

　　参见我的博文：全排列的编码与解码：康托展开

代码：

class Solution {
public:
    string getPermutation(int n, int k) {
        vector<int> temp;
        vector<int> fac;
        vector<int> nums;

        int next_k,now_seq;
        string result;

        if( n== ){
            result="";
            return result;
        }

        //generate nums
        for( int i=;i<n;i++ )
            nums.push_back(i+);
        //calculate factorial times,fac[0]=(n-1)!,fac[1]=(n-2)!
        for( int i=n-;i>=;i-- ){
            if( i==n- )
                fac.push_back();
            else
                fac.insert(fac.begin(),(n-i-)*(*fac.begin()));
        }
        //calculate every place from high
        next_k=k;
        for( int i=;i<n-;i++ ){
            now_seq=(next_k-)/fac[i]+;
            temp.push_back(nums[now_seq-]);
            nums.erase(nums.begin()+now_seq-);
            next_k=(next_k-)%fac[i]+;
        }
        temp.push_back(nums[]);
        //into string
        for( int i=;i<n;i++ ){
            result.append(,''+temp[i]);
        }
        return result;
    }
};