不知道写了多久。TestCase自己有问题：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 ArrayList<ArrayList<Integer>> combinationSum(int[] candidates, int target) {

        Arrays.sort(candidates);

        ArrayList<ArrayList<Integer>> result = new ArrayList<ArrayList<Integer>>();

        

        Set<ArrayList<Integer>> s = new HashSet<ArrayList<Integer>>();

        addResult(s, candidates, target, null);

        

        result.addAll(s);

        return result;

    }

    

    private void addResult(Set<ArrayList<Integer>> result, int[] candidate, int target, ArrayList<Integer> extras) {

        for (int i = 0; i < candidate.length && candidate[i] <= target; i++) {

            if (candidate[i] == target) {

                ArrayList<Integer> is = new ArrayList<Integer>();

                is.add(candidate[i]);

                if (extras != null && extras.isEmpty() == false) {

                    is.addAll(extras);

                    Collections.sort(is);

                }

                result.add(is);

            } else if (candidate[i] < target) {

                ArrayList<Integer> extras2 = null;

                if (extras == null) {

                    extras2 = new ArrayList<Integer>();

                } else {

                    extras2 = (ArrayList<Integer>) extras.clone();

                }

                extras2.add(candidate[i]);

                addResult(result, candidate, target - candidate[i], extras2);

            }

        }

    }

}