(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]
]

這道題我是這麼想的，假設結果集裏有一個2，那麼剩下的問題就變成了子問題，候選數集變成了[3,6,7]，target變成了5，從而可以考慮遞歸求解，若子問題返回空List，則說明不存在只有一個2的解，那麼疊加2，直到疊加的和大於target爲止。若等於target，也沒後面數的事了，可以返回了。

還有就是不需要原候選數是有序的

要說分類，可以歸爲遞歸，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;
	}
}