一、题目
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] ] 题目大意:意思说 给一组正数C,然后 给你一个目标数T, 让你从那组C中找到加在一起等于T的那些组合。 比如 给定target=7,然后 从[2,3,6,7]中可以找到[2,2,3]和[7]两组组合。
二、思路
用深度优先(DFS),看到题目后,有想到使用DFS思想,但是具体不太会实现,因为仅仅毕竟不太熟悉,看了别人的博客(后附),觉得写得很好,所以搬过来,供大家参考。
主要思想是:
Target =T,然后从数组中找一个数n,然后在 剩下的部分target 变成了 T-n,以此类推。函数到哪返回呢,如果目标数T=0,则找的成功,返回,如果目标数T小于C中最小的数,言外之意就是我们找到不这样的组合了,寻找失败,返回。 需要注意的是,答案要求没有重复的,如果只是这么写会变成[2,3,2],[2,2,3],[3,2,2],因此要记下 上一个数,我是从小往大找的,也就是说,如果我已经找完n=2的情况,再去找n=3的时候,3就不应该往回再选n=2了,只能往后走,不然就会重复。
三、代码
#coding:utf-8 class Solution: def combinationSum(self, candidates, target): """ :type candidates: List[int] :type target: int :rtype: List[List[int]] """ self.resList = [] candidates = sorted(candidates) self.dfs(candidates,[],target,0) print(self.resList) return self.resList def dfs(self,candidates,sublist,target,last): if target == 0: self.resList.append(sublist[:]) if target < candidates[0]: return for n in candidates: if n > target: return if n < last: continue sublist.append(n) self.dfs(candidates,sublist,target-n,n) sublist.pop() # def onlyonecandidate(self,candidate,target): # solutionset=[] # solutionsets = [] # if target % candidate == 0: # repeat = target // candidate # solutionset = [repeat for i in range(repeat)] #在solutionset中将repeat复制repeat次 # print(solutionset) # solutionsets.append(solutionset) # print(solutionsets) # return solutionsets if __name__ == ‘__main__‘: # repeat = [1,4,6] # for i in repeat: # print(i) candidates = [2,4,3,1] target = 5 ss = Solution() ss.combinationSum(candidates,target)
参考博客:https://blog.csdn.net/zl87758539/article/details/51693179
原文地址:https://www.cnblogs.com/xiaodongsuibi/p/8985750.html