Skip to content

Navigation Menu

Sign in
Appearance settings

Search code, repositories, users, issues, pull requests...

Provide feedback

We read every piece of feedback, and take your input very seriously.

Saved searches

Use saved searches to filter your results more quickly

Sign up
Appearance settings

[pull] master from youngyangyang04:master #275

New issue

Have a question about this project? Sign up for a free GitHub account to open an issue and contact its maintainers and the community.

By clicking “Sign up for GitHub”, you agree to our terms of service and privacy statement. We’ll occasionally send you account related emails.

Already on GitHub? Sign in to your account

Merged
pull merged 24 commits into AlgorithmAndLeetCode:master from youngyangyang04:master
May 29, 2023
Merged
Changes from 1 commit
Commits
Show all changes
24 commits
Select commit Hold shift + click to select a range
2f86a5c
Update 0701.二叉搜索树中的插入操作.md
jianghongcheng May 25, 2023
343e077
Update 0216.组合总和III.md
jianghongcheng May 26, 2023
89e9d75
Update 0077.组合.md
jianghongcheng May 26, 2023
4d48c45
Update 0077.组合优化.md
jianghongcheng May 26, 2023
847c60a
Update 0017.电话号码的字母组合.md
jianghongcheng May 26, 2023
2c51420
Update 0039.组合总和.md
jianghongcheng May 26, 2023
20db57f
Update 0040.组合总和II.md
jianghongcheng May 26, 2023
59742e5
Update 0040.组合总和II.md
jianghongcheng May 26, 2023
4586386
Update 0131.分割回文串.md
jianghongcheng May 27, 2023
78445d7
Update 0131.分割回文串.md
jianghongcheng May 27, 2023
51c8976
Update 0093.复原IP地址.md
jianghongcheng May 27, 2023
d45c141
Update 0078.子集.md
jianghongcheng May 27, 2023
5801ae7
Update 0090.子集II.md
jianghongcheng May 27, 2023
eb72f26
Update 0491.递增子序列.md
jianghongcheng May 28, 2023
7e3fc8e
Update 0491.递增子序列.md
jianghongcheng May 28, 2023
a6c95b7
Update 0491.递增子序列.md
jianghongcheng May 28, 2023
b033a08
Update 0046.全排列.md
jianghongcheng May 28, 2023
fd2608a
Update 0047.全排列II.md
jianghongcheng May 28, 2023
19ecde0
Update 回溯算法去重问题的另一种写法.md
jianghongcheng May 28, 2023
498395d
Update 0332.重新安排行程.md
jianghongcheng May 28, 2023
93e0a18
Update 0051.N皇后.md
jianghongcheng May 28, 2023
11581df
Update 0455.分发饼干.md
jianghongcheng May 28, 2023
18a406e
Update 0376.摆动序列.md
jianghongcheng May 28, 2023
996ddbe
Merge pull request #2098 from jianghongcheng/master
youngyangyang04 May 29, 2023
File filter

Filter by extension

Filter by extension

Conversations
Failed to load comments.
Loading
Jump to
Jump to file
Failed to load files.
Loading
Diff view
Diff view
Prev Previous commit
Next Next commit
Update 0046.全排列.md
  • Loading branch information
jianghongcheng authored May 28, 2023
commit b033a08c13547a43f3ac4a873f65151987569e6a
73 changes: 16 additions & 57 deletions problems/0046.全排列.md
View file Open in desktop
Original file line number Diff line number Diff line change
Expand Up @@ -213,68 +213,27 @@ class Solution {
```

### Python
**回溯**
回溯 使用used
```python
class Solution:
def __init__(self):
self.path = []
self.paths = []

def permute(self, nums: List[int]) -> List[List[int]]:
'''
因为本题排列是有序的,这意味着同一层的元素可以重复使用,但同一树枝上不能重复使用(usage_list)
所以处理排列问题每层都需要从头搜索,故不再使用start_index
'''
usage_list = [False] * len(nums)
self.backtracking(nums, usage_list)
return self.paths

def backtracking(self, nums: List[int], usage_list: List[bool]) -> None:
# Base Case本题求叶子节点
if len(self.path) == len(nums):
self.paths.append(self.path[:])
def permute(self, nums):
result = []
self.backtracking(nums, [], [False] * len(nums), result)
return result

def backtracking(self, nums, path, used, result):
if len(path) == len(nums):
result.append(path[:])
return

# 单层递归逻辑
for i in range(0, len(nums)): # 从头开始搜索
# 若遇到self.path里已收录的元素,跳过
if usage_list[i] == True:
for i in range(len(nums)):
if used[i]:
continue
usage_list[i] = True
self.path.append(nums[i])
self.backtracking(nums, usage_list) # 纵向传递使用信息,去重
self.path.pop()
usage_list[i] = False
```
**回溯+丢掉usage_list**
```python
class Solution:
def __init__(self):
self.path = []
self.paths = []

def permute(self, nums: List[int]) -> List[List[int]]:
'''
因为本题排列是有序的,这意味着同一层的元素可以重复使用,但同一树枝上不能重复使用
所以处理排列问题每层都需要从头搜索,故不再使用start_index
'''
self.backtracking(nums)
return self.paths

def backtracking(self, nums: List[int]) -> None:
# Base Case本题求叶子节点
if len(self.path) == len(nums):
self.paths.append(self.path[:])
return
used[i] = True
path.append(nums[i])
self.backtracking(nums, path, used, result)
path.pop()
used[i] = False

# 单层递归逻辑
for i in range(0, len(nums)): # 从头开始搜索
# 若遇到self.path里已收录的元素,跳过
if nums[i] in self.path:
continue
self.path.append(nums[i])
self.backtracking(nums)
self.path.pop()
```

### Go
Expand Down

AltStyle によって変換されたページ (->オリジナル) /