LeetCode_OJ【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.
• Elements in a combination (a₁, a₂, … , a_k) must be in non-descending order. (ie, a₁ ≤ a₂ ≤ … ≤ a_k).
• 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]

给定数组和目标值，从数组中找到所有的组合，使得组合的总和等于目标值，数组中的值可以重复选取。

这个题目的主要思路就是递归回溯，说来惭愧，好久没写递归，答案还是参考了别人的。

下面是本题的java实现。

public class Solution {
	public List<List<Integer>> combinationSum(int[] candidates, int target) {
		List<List<Integer>> res = new ArrayList<List<Integer>>();
		List<Integer> tmp = new ArrayList<Integer>();
		dfs(res,tmp,candidates,target,0);
		return res;
    }
	
	public void dfs(List<List<Integer>> res,List<Integer> tmp,int[] candidates, int gap,int index){
		if(gap == 0){
			List<Integer> list = new ArrayList<Integer>(tmp); 
			res.add(list);
			return;
		}
		for(int i = index ; i < candidates.length; i++){
			if(gap < candidates[index])
				return;
			else{
				tmp.add(candidates[i]);
				gap -= candidates[i];
				dfs(res,tmp,candidates,gap,i);
				tmp.remove(tmp.size()-1);
			}
		}
	}
}