Combination Sum.txt

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, … , 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] 

public class Solution{
    public List<List<Integer>> combinationSum(int[] candidates, int target){
        
    }
}

My first attempt: (the code is wrong). 
public class Solution{
    public ArrayList<ArrayList<Integer>> combinationSum(int[] candidates, int target) {
        ArrayList<ArrayList<Integer>> result = new ArrayList<ArrayList<Integer>>();
        Arrays.sort(candidates);
        ArrayList<Integer> item = new ArrayList<Integer>();
        dfs(candidates, item, target, result);
        return result;
    }
    public void dfs(int[] candidates, ArrayList<Integer> item, int target, ArrayList<ArrayList<Integer>> result){
        if(target == 0){
            result.add(new ArrayList<Integer>(item));
        }
        for(int i = 0; i < candidates.length; i++){
            item.add(candidates[i]);
            dfs(candidates, item, target-candidates[i], result);
            item.remove(item.size()-1);
        }
    }
}
Correct Solution: 
Time: O(n!), Space: O(n)
public class Solution {
    public ArrayList<ArrayList<Integer>> combinationSum(int[] candidates, int target) {
        ArrayList<ArrayList<Integer>> result = new ArrayList<ArrayList<Integer>>();
        Arrays.sort(candidates);
        ArrayList<Integer> item = new ArrayList<Integer>();
        dfs(candidates, 0, item, target, result);
        return result;
    }
    public void dfs(int[] candidates, int start, ArrayList<Integer> item, int target, ArrayList<ArrayList<Integer>> result){
        if(target == 0){
            result.add(new ArrayList<Integer>(item));
        }
        for(int i = start; i < candidates.length; i++){
            if(target < candidates[i]) return;
            if(i > 0 && candidates[i] == candidates[i-1]) continue;//因为每个数本身就可以重复取，所以数组里的相同数可以跳过。
            item.add(candidates[i]);
            dfs(candidates, i, item, target-candidates[i], result);//dfs里的参数还是i,而不是i+1是因为每个数可以重复取。
            item.remove(item.size()-1);
        }
    }
}

I think adding this line can cut off paths in dfs search. 
if(target < candidates[i]) return;
But if we don't have this line, we need to check if(target < 0):
public void dfs(int[] candidates, int start, ArrayList<Integer> item, int target, ArrayList<ArrayList<Integer>> result){
	if(target < 0) return;
	if(target == 0){
		result.add(new ArrayList<Integer>(item));
	}
	for(int i = start; i < candidates.length; i++){
		//if(target < candidates[i]) return;
		if(i > 0 && candidates[i] == candidates[i-1]) continue;
		item.add(candidates[i]);
		dfs(candidates, i, item, target-candidates[i], result);
		item.remove(item.size()-1);
	}
}