39. Combination Sum（dfs）

最新推荐文章于 2021-05-13 18:54:27 发布

转载最新推荐文章于 2021-05-13 18:54:27 发布 · 77 阅读

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原文链接：http://www.cnblogs.com/bernieloveslife/p/9784039.html

本文深入探讨了组合求和算法，旨在寻找所有可能的组合，使得候选数的总和等于目标数。允许从候选集中重复选取元素。文章通过具体实例展示了如何使用深度优先搜索(DFS)来解决这一问题，确保解决方案集合中没有重复的组合。

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]]

class Solution:
    def combinationSum(self, candidates, target):
        """
        :type candidates: List[int]
        :type target: int
        :rtype: List[List[int]]
        """
        res = []
        candidates.sort()
        def dfs(i,target,s):
            if i>=len(candidates):
                return
            if candidates[i]==target:
                res.append(s+[candidates[i]])
                return
            if target<=0:
                return
            dfs(i+1,target,s)
            dfs(i,target-candidates[i],s+[candidates[i]])
        dfs(0,target,[])
        return res

转载于:https://www.cnblogs.com/bernieloveslife/p/9784039.html