LeetCode解题报告--Combination Sum

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]

来源：CSDN

作者：YatKam

链接：https://blog.csdn.net/github_27609763/article/details/50210539