LeetCode解题报告--Combination Sum
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题目:
Given a set of candidate numbers (C) and a target number (T), find all unique combinations in C where the candidate numbers sums to T.
The same repeated number may be chosen from C unlimited number of times.
Note:
All numbers (including target) will be positive integers.
Elements in a combination (a1, a2, … , ak) must be in non-descending order. (ie, a1 ≤ a2 ≤ … ≤ ak).
The solution set must not contain duplicate combinations.
For example, given candidate set 2,3,6,7 and target 7,
A solution set is:
[7]
[2, 2, 3]
题目链接:https://leetcode.com/problems/combination-sum/
分析:题意给定一组数组,找出所有满足组合的元素和=targetDFS数组元素可重复使用。直接用递归求解,与之前LeetCode的3sum的有相似之处,可参考一下。
将给定数组排列之后,进行递归,用(target - candidates【i】(即当前元素))作为下次递归的target,每次的递归从0–candidates.length,当target<0直接跳出递归,target == 0时即为符合要求的组合。注意去掉重复元素产生的重复组合。
具体递归演示如图示:
Java代码 Accepted:
public class Solution { public List<List<Integer>> combinationSum(int[] candidates, int target) { List<Integer> subRes = new ArrayList<Integer>(); List<List<Integer>> result = new ArrayList<List<Integer>>(); //Special case if(candidates.length == 0 || candidates == null) return result; //Sort array before find combination Arrays.sort(candidates); //Recursive DFS recursionFunction(candidates,0,target,subRes,result); return result; } public static void recursionFunction(int[] candidates,int begin,int target,List<Integer> subRes,List<List<Integer>> result){ if(target < 0) return; else if(target == 0){ //Have to redeclarat subRes as ArrayList<Integer> or can't get resSub result.add(new ArrayList<Integer>(subRes)); return; }else{ for(int i = begin;i < candidates.length;i ++){ //To eliminate repeate elment of candidates if(i > 0 && candidates[i] == candidates[i - 1]) continue; subRes.add(candidates[i]); recursionFunction(candidates,i,target - candidates[i],subRes,result); subRes.remove(subRes.size() - 1); } } }}
Python代码 Accepted
class Solution(object): def __init__(self): self.res = [] self.subRes = [] self.target = 0 def combinationSum(self, candidates, target): """ :type candidates: List[int] :type target: int :rtype: List[List[int]] """ if(len(candidates) == 0 or candidates == None): return self.res candidates = sorted(candidates) self.recursionFunction(candidates,0,target,self.subRes,self.res) return self.res def recursionFunction(self,candidates,begin,target,subRes,res): if(target < 0): return elif(target == 0): self.res.append(self.subRes) return else: for i in range(begin,len(candidates)): if(i > 0 and candidates[i] == candidates[i - 1]): continue self.subRes.append(candidates[i]) self.recursionFunction(candidates,i,target - candidates[i],self.subRes,self.res) self.subRes = self.subRes[:-1]
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