[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):

"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.

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;
}
};