Update 0039.组合总和.md

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jianghongcheng
2023-05-26 10:28:01 -05:00
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commit 2c51420308

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@ -271,75 +271,101 @@ class Solution {
## Python
**回溯**
回溯(版本一)
```python
class Solution:
def __init__(self):
self.path = []
self.paths = []
def combinationSum(self, candidates: List[int], target: int) -> List[List[int]]:
'''
因为本题没有组合数量限制所以只要元素总和大于target就算结束
'''
self.path.clear()
self.paths.clear()
self.backtracking(candidates, target, 0, 0)
return self.paths
def backtracking(self, candidates: List[int], target: int, sum_: int, start_index: int) -> None:
# Base Case
if sum_ == target:
self.paths.append(self.path[:]) # 因为是shallow copy所以不能直接传入self.path
def backtracking(self, candidates, target, total, startIndex, path, result):
if total > target:
return
if sum_ > target:
if total == target:
result.append(path[:])
return
# 单层递归逻辑
for i in range(start_index, len(candidates)):
sum_ += candidates[i]
self.path.append(candidates[i])
self.backtracking(candidates, target, sum_, i) # 因为无限制重复选取所以不是i+1
sum_ -= candidates[i] # 回溯
self.path.pop() # 回溯
for i in range(startIndex, len(candidates)):
total += candidates[i]
path.append(candidates[i])
self.backtracking(candidates, target, total, i, path, result) # 不用i+1了表示可以重复读取当前的数
total -= candidates[i]
path.pop()
def combinationSum(self, candidates, target):
result = []
self.backtracking(candidates, target, 0, 0, [], result)
return result
```
**剪枝回溯**
回溯剪枝(版本一)
```python
class Solution:
def __init__(self):
self.path = []
self.paths = []
def combinationSum(self, candidates: List[int], target: int) -> List[List[int]]:
'''
因为本题没有组合数量限制所以只要元素总和大于target就算结束
'''
self.path.clear()
self.paths.clear()
# 为了剪枝需要提前进行排序
candidates.sort()
self.backtracking(candidates, target, 0, 0)
return self.paths
def backtracking(self, candidates: List[int], target: int, sum_: int, start_index: int) -> None:
# Base Case
if sum_ == target:
self.paths.append(self.path[:]) # 因为是shallow copy所以不能直接传入self.path
def backtracking(self, candidates, target, total, startIndex, path, result):
if total == target:
result.append(path[:])
return
# 单层递归逻辑
# 如果本层 sum + condidates[i] > target就提前结束遍历剪枝
for i in range(start_index, len(candidates)):
if sum_ + candidates[i] > target:
return
sum_ += candidates[i]
self.path.append(candidates[i])
self.backtracking(candidates, target, sum_, i) # 因为无限制重复选取所以不是i-1
sum_ -= candidates[i] # 回溯
self.path.pop() # 回溯
for i in range(startIndex, len(candidates)):
if total + candidates[i] > target:
break
total += candidates[i]
path.append(candidates[i])
self.backtracking(candidates, target, total, i, path, result)
total -= candidates[i]
path.pop()
def combinationSum(self, candidates, target):
result = []
candidates.sort() # 需要排序
self.backtracking(candidates, target, 0, 0, [], result)
return result
```
回溯(版本二)
```python
class Solution:
def combinationSum(self, candidates: List[int], target: int) -> List[List[int]]:
result =[]
self.backtracking(candidates, target, 0, [], result)
return result
def backtracking(self, candidates, target, startIndex, path, result):
if target == 0:
result.append(path[:])
return
if target < 0:
return
for i in range(startIndex, len(candidates)):
path.append(candidates[i])
self.backtracking(candidates, target - candidates[i], i, path, result)
path.pop()
```
回溯剪枝(版本二)
```python
class Solution:
def combinationSum(self, candidates: List[int], target: int) -> List[List[int]]:
result =[]
candidates.sort()
self.backtracking(candidates, target, 0, [], result)
return result
def backtracking(self, candidates, target, startIndex, path, result):
if target == 0:
result.append(path[:])
return
for i in range(startIndex, len(candidates)):
if target - candidates[i] < 0:
break
path.append(candidates[i])
self.backtracking(candidates, target - candidates[i], i, path, result)
path.pop()
```
## Go