[leetcode] 39. Combination Sum @ python

目标组合求解

最新推荐文章于 2022-11-23 12:18:42 发布

原创最新推荐文章于 2022-11-23 12:18:42 发布 · 486 阅读

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本文介绍了一种使用深度优先搜索(DFS)和回溯法解决给定候选数集合和目标数，寻找所有可能组合的问题。通过遍历候选数，递归地寻找所有可能的组合，直到找到目标数的组合。

原题

Given a set of candidate numbers (candidates) (without duplicates) and a target number (target), find all unique combinations in candidates where the candidate numbers sums to target.

The same repeated number may be chosen from candidates unlimited number of times.

Note:

All numbers (including target) will be positive integers.
The solution set must not contain duplicate combinations.
Example 1:

Input: candidates = [2,3,6,7], target = 7,
A solution set is:
[
[7],
[2,2,3]
]
Example 2:

Input: candidates = [2,3,5], target = 8,
A solution set is:
[
[2,2,2,2],
[2,3,3],
[3,5]
]

解法

DFS + backtracking(回溯法), 回溯的条件是当target为负时, 返回; 当target为0时, 说明我们找到了匹配的组合, 因此将path加到res并返回; DFS的主体是遍历candidates, 递归找出所有可能的组合.

代码

class Solution:
    def combinationSum(self, candidates, target):
        """
        :type candidates: List[int]
        :type target: int
        :rtype: List[List[int]]
        """
        candidates.sort()
        res = []
        self.dfs(candidates, target, 0, [], res)
        return res
        
    def dfs(self, nums, target, index, path, res):
        # backtracking condition
        if target < 0:
            return
        if target == 0:
            # we find a match
            res.append(path)
            return
        for i in range(index, len(nums)):
            self.dfs(nums, target-nums[i], i, path + [nums[i]], res)