[leedcode 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, … , 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 {
    List<Integer> seq;
    List<List<Integer>> res;
    public List<List<Integer>> combinationSum(int[] candidates, int target) {
        //类似全排列的变形，首先找到全排列，for循环加递归，终止条件是如果sum>target直接返回，如果sum=target则保存结果
        //注意for循环的开始条件，以及递归的变量，本题允许每个数使用多次。level代表递归开始的candidates下标
        //非常注意：：在res中填值时，要重新new一个Integer，不能使用之前的，因为再操作seq会改变已经保存的值！
        Arrays.sort(candidates);
        seq=new ArrayList<Integer>();
        res=new ArrayList<List<Integer>>();
        find(candidates,target,0,0);

        return res;
    }
    public void find(int[] candidates,int target,int sum,int level){
        if(sum==target){
            res.add(new ArrayList<Integer>(seq));
            return;
        }
        if(sum>target){
            return;
        }
        for(int i=level;i<candidates.length;i++){
            seq.add(candidates[i]);
            sum+=candidates[i];
            find(candidates,target,sum,i);
            seq.remove(seq.size()-1);//
            sum-=candidates[i];
        }
        
    }
}