[Leetcode] 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] 

Solution:

public class Solution {
    public List<List<Integer>> combinationSum(int[] candidates, int target) {
        List<List<Integer>> result=new ArrayList<List<Integer>>();
        List<Integer> al=new ArrayList<Integer>();
        Arrays.sort(candidates);
        dfs(result,al,target,candidates,0);
        return result;
    }

    private void dfs(List<List<Integer>> result, List<Integer> al, int target, int[] candidates, int position) {
        // TODO Auto-generated method stub
        if(target<0)
            return;
        if(target==0){
            result.add(new ArrayList<Integer>(al));
            return;
        }
        for(int i=position;i<candidates.length;++i){
            al.add(candidates[i]);
            dfs(result, al, target-candidates[i], candidates, i);
            al.remove(al.size()-1);
        }    
    }
}

注意：

这里在循环的dfs前al里加了candidates[i],在dfs出来后，得remove掉，但是注意啊注意,这里得依次删掉最后一个元素。

dfs中的参数position是i，不能继续用position了！！