leetcode: 60. Permutation Sequence

Problem

# 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.

Idea

先通过举例来获得更好的理解。以n = 4，k = 9为例：

1234
1243
1324
1342
1423
1432
2134
2143
2314  <= k = 9
2341
2413
2431
3124
3142
3214
3241
3412
3421
4123
4132
4213
4231
4312
4321

最高位可以取{1, 2, 3, 4}，而每个数重复3! = 6次。所以第k=9个permutation的s[0]为{1, 2, 3, 4}中的第9/6+1 = 2个数字s[0] = 2。

而对于以2开头的6个数字而言，k = 9是其中的第k' = 9%(3!) = 3个。而剩下的数字{1, 3, 4}的重复周期为2! = 2次。所以s[1]为{1, 3, 4}中的第k'/(2!)+1 = 2个，即s[1] = 3。

对于以23开头的2个数字而言，k = 9是其中的第k'' = k'%(2!) = 1个。剩下的数字{1, 4}的重复周期为1! = 1次。所以s[2] = 1.

对于以231开头的一个数字而言，k = 9是其中的第k''' = k''/(1!)+1 = 1个。s[3] = 4

AC

class Solution():
def getPermutation(self, n, k):
import math
seq, k, fact, perm = "", k - 1, math.factorial(n - 1), list(range(1, n + 1))
for i in reversed(range(n)):
curr = perm[k // fact]
seq += str(curr)
perm.remove(curr)
if i > 0:
k %= fact
fact //= i
return seq

if __name__ == "__main__":
assert Solution().getPermutation(3, 2) == '132'