# LeetCode | 0040. Combination Sum II组合总和 II【Python】

### Problem

LeetCode

Given a collection of candidate numbers (`candidates`) and a target number (`target`), find all unique combinations in `candidates` where the candidate numbers sums to `target`.

Each number in `candidates` may only be used once in the combination.

Note:

• All numbers (including `target`) will be positive integers.
• The solution set must not contain duplicate combinations.

Example 1:

``````Input: candidates = [10,1,2,7,6,1,5], target = 8,
A solution set is:
[
[1, 7],
[1, 2, 5],
[2, 6],
[1, 1, 6]
]
``````

Example 2:

``````Input: candidates = [2,5,2,1,2], target = 5,
A solution set is:
[
[1,2,2],
[5]
]
``````

### 问题

candidates 中的每个数字在每个组合中只能使用一次。

• 所有数字（包括目标数）都是正整数。
• 解集不能包含重复的组合。

``````输入: candidates = [10,1,2,7,6,1,5], target = 8,

[
[1, 7],
[1, 2, 5],
[2, 6],
[1, 1, 6]
]
``````

``````输入: candidates = [2,5,2,1,2], target = 5,

[
[1,2,2],
[5]
]
``````

### 思路

``````res = []
def backtrack(路径, 选择列表):
if 满足结束条件:
res.append(路径)
return

for 选择 in 选择列表:
做选择
backtrack(路径, 选择列表)
撤销选择
``````
##### Python3 代码
``````from typing import List

class Solution:
def combinationSum2(self, candidates: List[int], target: int) -> List[List[int]]:
n = len(candidates)
if n == 0:
return []

# accelerate 剪枝提速，非必需
candidates.sort()

path, res = [], []
self.dfs(candidates, 0, n, path, res, target)
return res

def dfs(self, candidates, start, n, path, res, target):
# 1.valid result 递归终止情况
if target == 0:
res.append(path[:])
return

for i in range(start, n):
tmp = target - candidates[i]
# 3.pruning 剪枝
if tmp < 0:
break
# 防止出现这种情况：一个数字使用多次
if i > start and candidates[i] == candidates[i - 1]:
continue
# 2.backtrack and update 回溯以及更新 path
path.append(candidates[i])
self.dfs(candidates, i + 1, n, path, res, tmp)
path.pop()
``````

Python