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.
分析:
使用康托展开,直接生成第k个排列。了解原理请看:康托展开。
代码:
class Solution(object):
def getPermutation(self, n, k):
"""
:type n: int
:type k: int
:rtype: str
"""
remains = k - 1
div = []
for i in range(n):
a, remains = divmod(remains, self.getFactory(n - i - 1))
div.append(a)
res = ''
seq = range(1, n + 1)
for i in div:
res += str(seq[i])
del seq[i]
return res
def getFactory(self, n):
if n in [0, 1]:
return 1
return reduce(lambda x, y: x*y, range(1, n + 1))