60. Permutation Sequence

最新推荐文章于 2020-04-14 00:08:23 发布

转载最新推荐文章于 2020-04-14 00:08:23 发布 · 93 阅读

本文介绍了一种算法，用于找出数字集合[1,2,3,...,n]中第k个唯一排列。通过列举并标记所有可能的排列顺序，可以得到当n=3时的排列序列。该算法使用C++实现，能够处理n在1到9之间的输入，并确保k在1到n!的范围内。

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:

"123"
"132"
"213"
"231"
"312"
"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"

class Solution {
public:
    string getPermutation(int n, int k) {
        if(n<=0)
            return "";
        vector<int> num;
        for(int i=1; i<=n; i++)
            num.push_back(i);
        int factorial = 1;
        for(int i=2; i<n; i++)
            factorial *=i;
        int round = n-1;
        k--;
        string ret = "";
        while(round>=0)
        {
            int index = k/factorial;
            k%=factorial;
            ret+=to_string(num[index]);
            vector<int>::iterator iter=find(num.begin(), num.end(), num[index]);
            num.erase(iter);
            if(round>0)
                factorial/=round;
            round--;
        }
        return ret;
    }
};