[leetcode] 40. Combination Sum II @ python_leetcode 40 combination sum python-优快云博客

本文链接：https://blog.youkuaiyun.com/danspace1/article/details/88074092

本文探讨了如何使用深度优先搜索(DFS)和回溯法解决候选数集合中所有唯一组合的问题，这些组合的和等于目标数。在解决方案中，每个候选数只能使用一次，并且不允许出现重复的组合。文章提供了Python代码实现，通过排序和跳过重复元素来避免重复的组合。

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原题

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]
]

解法

DFS + 回溯法. 回溯的条件是target == 0时, 表示我们已经找到一个组合, 此时将path加到结果中. Base case是当nums长度为0且target不为0时, 直接return.然后遍历nums, 如果nums[i]==nums[i-1]时, 直接忽略,以此去重.

代码

class Solution(object):
    def combinationSum2(self, nums, target):
        """
        :type nums: List[int]
        :type target: int
        :rtype: List[List[int]]
        """
        def dfs(nums, target, path, res):
            if len(nums) == 0 and target != 0:
                return
            if target == 0:
                res.append(path)
                return
            for i in range(len(nums)):
                if i > 0 and nums[i] == nums[i-1]:
                    continue
                if nums[i] <= target:
                    dfs(nums[i+1:], target-nums[i], path+[nums[i]], res)               
             
        nums.sort()
        res = []
        dfs(nums, target, [], res)
        return res