Topic 46 Permutations I
Given a collection of distinct integers, return all possible permutations.
Example:
Input: [1,2,3]
Output:
[
[1,2,3],
[1,3,2],
[2,1,3],
[2,3,1],
[3,1,2],
[3,2,1]
]
Thinking
For the number of n, n-1 corresponds to the first number before the arrangement, this last number and then inserted into a different location
def permute(self, nums: List[int]) -> List[List[int]]:
if len(nums) == 0:
return []
elif len(nums) == 1:
return [[nums[0]]]
ans = []
temp = self.permute(nums[1:])
for item in temp:
for i in range(len(item)+1):
p = item.copy()
p.insert(i,nums[0])
ans.append(p)
return ans
Topic 47 Permutations II
Given a collection of numbers that might contain duplicates, return all possible unique permutations.
Example:
Input: [1,1,2]
Output:
[
[1,1,2],
[1,2,1],
[2,1,1]
]
Thinking
Recursion is also inserted, but at the time of each insert a new number, if you encounter the same number and insert a number, you can terminate. Because of the back of the room, at the time of the last time you insert a digital has been considered and, side by side in the current item.
def permuteUnique(self, nums: List[int]) -> List[List[int]]:
if len(nums) == 0:
return []
elif len(nums) == 1:
return [[nums[0]]]
nums.sort()
ans = []
temp = self.permuteUnique(nums[1:])
for item in temp:
for i in range(len(item)+1):
p = item.copy()
p.insert(i,nums[0])
ans.append(p)
if i<len(item):
if nums[0] == item[i]:
break
return ans
