Leetcode153: 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.

在n!个排列中，第一位的元素总是(n-1)!一组出现的，也就说如果p = k / (n-1)!，那么排列的最开始一个元素一定是nums[p]。

假设有n个元素，第K个permutation是

a1, a2, a3, ..... ..., an

那么a1是哪一个数字呢？

那么这里，我们把a1去掉，那么剩下的permutation为

a2, a3, .... .... an, 共计n-1个元素。 n-1个元素共有(n-1)!组排列，那么这里就可以知道

设变量K1 = K

a1 = K1 / (n-1)!

同理，a2的值可以推导为

a2 = K2 / (n-2)!

K2 = K1 % (n-1)!

.......

a(n-1) = K(n-1) / 1!

K(n-1) = K(n-2) /2!

an = K(n-1)

class Solution {
public:
string getPermutation(int n, int k) {
vector<int> nums(n);
int pCount = 1;
for(int i = 0 ; i < n; ++i)
{
nums[i] = i + 1;
pCount *= (i + 1);
}

k--;
string res = "";
for(int i = 0 ; i < n; i++)
{
pCount = pCount/(n-i);
int selected = k / pCount;
res += ('0' + nums[selected]);
for(int j = selected; j < n-i-1; j++)
nums[j] = nums[j+1];
k = k % pCount;
}
return res;
}
};