Leetcode:60. Permutation Sequence

Description

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):

"123"
"132"
"213"
"231"
"312"
"321"
Given n and k, return the kth permutation sequence.

思路

• 分析，n个数的排列一共有n!个，
• 其中分别以1,2,...,n-1开头的各有(n-1)!种
• 所以，index = (k - 1) / ((n-1)!)为第一个开头的数字
• 然后k = k - index * (n-1)! 为剩下的第几个
• 然后重复计算剩下的(n-1)个数字即可。

代码

class Solution {
public:
    string getPermutation(int n, int k) {
        string res;
        vector<char> vec;
        int sum = 1;
        for(int i = 0; i < n; ++i){
            vec.push_back('1' + i);
            sum *= (i + 1);
        }
        
        int index = 0;
        int i = 0, tmp = 0;
        while(n > 0){
            tmp = sum / n;
            index = (k - 1) / tmp;
            res += vec[index];
            k = k - index * tmp;
            sum /= n;
            n--;
        }
        
        return res;
    }
};