[leetcode] 60. Permutation Sequence 解题报告

题目链接：https://leetcode.com/problems/permutation-sequence/

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

Given n and k, return the kth permutation sequence.

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

思路：给定ｎ个数字让求第k个序列．ｎ个数字总共的全排列最多有n!个，并且全排列有个规律．

给定一个n，以１开头的排列有(n-1)!个，同样对２和３也是，所以如果k小于等于(n-1)!，那么首位必为１，因为以１开头的全排列有(n-1)!个．

同样如果k大于(n-1)!，那么第一个数就应该为(k-1)/(n-1)! + 1．这样找到了首位数字应该是哪个，剩下了(n-1)个数字，我们只需要再重复上面的步骤，不断缩小k即可．

代码如下：

class Solution {public:    string getPermutation(int n, int k) {        vector<int> sum(n, 1), index(n, 1);        string result;        for(int i = 1; i < n; i++)        {            sum[i] = sum[i-1]*i;            index[i] = i+1;        }        k--;        for(int i = n-1; i >=0; i--)        {            int pos = k/sum[i];            result += to_string(index[pos]);            index.erase(index.begin() + pos);             k = k%sum[i];        }        return result;    }};