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, … , 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]

题解：

还是老问题，用DFS找解决方案，不同点是，这道题： The same repeated number may be chosen from C unlimited number of times.

所以，每次跳进递归不用都往后挪一个，还可以利用当前的元素尝试。

同时，这道题还要判断重复解。用我之前介绍的两个方法：

 1.       if(i>0 && candidates[i] == candidates[i-1])//deal with dupicate
                 continue; 

 2.       if(!res.contains(item)) 
                res.add(new ArrayList<Integer>(item));   

这两个方法解决。 

代码如下：

public ArrayList<ArrayList<Integer>> combinationSum(int[] candidates, int target) {  
        ArrayList<ArrayList<Integer>> res = new ArrayList<ArrayList<Integer>>();  
        ArrayList<Integer> item = new ArrayList<Integer>();
        if(candidates == null || candidates.length==0)  
            return res; 
            
        Arrays.sort(candidates);  
        helper(candidates,target, 0, item ,res);  
        return res;  
    }  
    
    private void helper(int[] candidates, int target, int start, ArrayList<Integer> item,   
    ArrayList<ArrayList<Integer>> res){  
        if(target<0)  
            return;  
        if(target==0){  
            res.add(new ArrayList<Integer>(item));  
            return;  
        }
        
        for(int i=start;i<candidates.length;i++){  
            if(i>0 && candidates[i] == candidates[i-1])//deal with dupicate
                continue;  
            item.add(candidates[i]);
            int newtarget = target - candidates[i];
            helper(candidates,newtarget,i,item,res);//之所以不传i+1的原因是：
                                                    //The same repeated number may be 
                                                    //chosen from C unlimited number of times.
            item.remove(item.size()-1);  
        }  
    }

reference: http://www.cnblogs.com/springfor/p/3884294.html