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

缺乏、安全感 2022-12-05 12:08 82阅读 0赞

### 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 ###

[力扣][Link 1]

给定一个数组 candidates 和一个目标数 target ，找出 candidates 中所有可以使数字和为 target 的组合。

candidates 中的每个数字在每个组合中只能使用一次。

**说明：**

*  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。

**示例 1:**

输入: candidates = [10,1,2,7,6,1,5], target = 8,
    所求解集为:
    [
      [1, 7],
      [1, 2, 5],
      [2, 6],
      [1, 1, 6]
    ]

**示例 2:**

输入: candidates = [2,5,2,1,2], target = 5,
    所求解集为:
    [
      [1,2,2],
      [5]
    ]

### 思路 ###

**回溯模板**

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()

### GitHub 链接 ###

[Python][]

[LeetCode]: https://leetcode.com/problems/combination-sum-ii/
[Link 1]: https://leetcode-cn.com/problems/combination-sum-ii/
[Python]: https://github.com/Wonz5130/LeetCode-Solutions/blob/master/solutions/0040-Combination-Sum-II/0040.py