难度: Medium
原题连接
内容描述
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 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.
Given k will be between 1 and n! inclusive.
Example 1:
Input: n = 3, k = 3
Output: "213"
Example 2:
Input: n = 4, k = 9
Output: "2314"
思路 1
当然是暴力直接算出所有的排列然后取第k个，但是会超时
class Solution(object):
def getPermutation(self, n, k):
"""
:type n: int
:type k: int
:rtype: str
"""
s = ''.join([str(i) for i in range(1, n+1)])
print(s)
def permunation(s):
if len(s) == 0:
return
if len(s) == 1:
return [s]
res = []
for i in range(len(s)):
x = s[i]
xs = s[:i] + s[i+1:]
for j in permunation(xs):
res.append(x+j)
return res
return permunation(s)[k-1]
思路 2
参考大神tso的思路
I'm sure somewhere can be simplified so it'd be nice if anyone can let me know. The pattern was that:
say n = 4, you have {1, 2, 3, 4}
If you were to list out all the permutations you have
1 + (permutations of 2, 3, 4)
2 + (permutations of 1, 3, 4)
3 + (permutations of 1, 2, 4)
4 + (permutations of 1, 2, 3)
We know how to calculate the number of permutations of n numbers... n! So each of those with permutations of 3 numbers means there are 6 possible permutations. Meaning there would be a total of 24 permutations in this particular one. So if you were to look for the (k = 14) 14th permutation, it would be in the
3 + (permutations of 1, 2, 4) subset.
To programmatically get that, you take k = 13 (subtract 1 because of things always starting at 0) and divide that by the 6 we got from the factorial, which would give you the index of the number you want. In the array {1, 2, 3, 4}, k/(n-1)! = 13/(4-1)! = 13/3! = 13/6 = 2. The array {1, 2, 3, 4} has a value of 3 at index 2. So the first number is a 3.
Then the problem repeats with less numbers.
The permutations of {1, 2, 4} would be:
1 + (permutations of 2, 4)
2 + (permutations of 1, 4)
4 + (permutations of 1, 2)
But our k is no longer the 14th, because in the previous step, we've already eliminated the 12 4-number permutations starting with 1 and 2. So you subtract 12 from k.. which gives you 1. Programmatically that would be...
k = k - (index from previous) * (n-1)! = k - 2*(n-1)! = 13 - 2*(3)! = 1
In this second step, permutations of 2 numbers has only 2 possibilities, meaning each of the three permutations listed above a has two possibilities, giving a total of 6. We're looking for the first one, so that would be in the 1 + (permutations of 2, 4) subset.
Meaning: index to get number from is k / (n - 2)! = 1 / (4-2)! = 1 / 2! = 0.. from {1, 2, 4}, index 0 is 1
so the numbers we have so far is 3, 1... and then repeating without explanations.
{2, 4}
k = k - (index from pervious) * (n-2)! = k - 0 * (n - 2)! = 1 - 0 = 1;
third number's index = k / (n - 3)! = 1 / (4-3)! = 1/ 1! = 1... from {2, 4}, index 1 has 4
Third number is 4
{2}
k = k - (index from pervious) * (n - 3)! = k - 1 * (4 - 3)! = 1 - 1 = 0;
third number's index = k / (n - 4)! = 0 / (4-4)! = 0/ 1 = 0... from {2}, index 0 has 2
Fourth number is 2
Giving us 3142. If you manually list out the permutations using DFS method, it would be 3142. Done! It really was all about pattern finding.
class Solution(object):
def getPermutation(self, n, k):
"""
:type n: int
:type k: int
:rtype: str
"""
res = ''
factorial = [1] * (n+1)
# factorial[] = [1, 1, 2, 6, 24, ... n!]
for i in range(1, n+1):
factorial[i] = factorial[i-1] * i
# create a list of numbers to get indices
nums = [i for i in range(1, n+1)]
# because we start from index 0
k -= 1
for i in range(1, n+1):
# this is the idx of first num each time we will get
idx = k / factorial[n-i]
res += str(nums[idx])
# delete this num, since we have got it
nums.pop(idx)
# update k
k -= idx * factorial[n-i]
return res或者空间复杂度更低一点
class Solution(object):
def getPermutation(self, n, k):
"""
:type n: int
:type k: int
:rtype: str
"""
seq, k, fact = '', k-1, math.factorial(n-1)
perm = [i for i in range(1, n+1)]
for i in range(n)[::-1]:
curr = perm[k/fact]
seq += str(curr)
perm.remove(curr)
if i > 0:
k %= fact
fact /= i
return seq