(Java)LeetCode-39. Combination Sum

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.
• 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]]

这道题我是这么想的，假设结果集里有一个2，那么剩下的问题就变成了子问题，候选数集变成了[3,6,7]，target变成了5，从而可以考虑递归求解，若子问题返回空List，则说明不存在只有一个2的解，那么叠加2，直到叠加的和大于target为止。若等于target，也没后面数的事了，可以返回了。

还有就是不需要原候选数是有序的

要说分类，可以归为递归，回溯，DFS吧

public class Solution {    public List<List<Integer>> combinationSum(int[] candidates, int target) {return combination(candidates, target, 0);    }public List<List<Integer>> combination(int[] candidates, int target, int index){List<List<Integer>> result = new ArrayList<List<Integer>>();for(int i = index; i < candidates.length; i++){List<Integer> list = new ArrayList<Integer>();int temp = candidates[i];while(temp <= target){list.add(candidates[i]);if(temp == target){result.add(list);break;}List<List<Integer>> tempResult = combination(candidates, target - temp, i + 1);for(List<Integer> res : tempResult){res.addAll(list);result.add(res);}temp += candidates[i];}}return result;}}