[Leetcode] Permutation sequnce

Original title address: https://oj.leetcode.com/submissions/detail/5341904/

Test instructions

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 would be between 1 and 9 inclusive.

Problem-solving ideas: Just started with DFS, but has been tle. It seems that Python is really slow to write Dfs in Java and C + +. Can only adopt a more ingenious way of thinking.

　　　　　

In fact, the data is not big, n up to 9,9! = 362880, the enumeration should be able to pass (I did not experiment).

The method I used was to calculate the K-permutation.

Suppose n = 6,k = 400

The first digit is calculated first,

The first bit is 6, so it's also the 5th at least! * 5 + 1 permutations, this is because when the first bit is 1/2/3/4/5, there are 5! permutations, so the first bit is 6 o'clock, at least the 5th! * 5 + 1 permutations (this arrangement is 612345).

5! * 5 + 1 = 601 > K, so the first bit cannot be 6.

Enumerate one by one until the first bit is 4 o'clock, at which point 4xxxxx is at least 5th! * 3 + 1 = 361 permutations.

Then the second digit is calculated,

The difference with the first digit is that the 46xxxx is at least 4th! * 4 + 1 = 97 permutations, this is because the smaller than 6 is only 5/3/2/1.

Finally, we can calculate the second bit as 2.

Finally, the No. 400 arrangement was 425361.

Code:

classSolution:#@return A string    defgetpermutation (self, n, k):#http://www.cnblogs.com/zuoyuan/p/3785530.htmlres ="'FAC= 1k-= 1 forIinchRange (1,n): FAC *=I num= [I forIinchRange (1,10)]         forIinchReversed (range (n)): Curr= num[k/FAC] Res+=Str (curr) num.remove (curr)ifI! =0:k%=FAC FAC/=IreturnRes

Reference:

Http://www.cnblogs.com/zuoyuan/p/3785530.html

[Leetcode] Permutation sequnce