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.

public class Solution {
    /** * @param candidates: A list of integers * @param target:An integer * @return: A list of lists of integers */
    public List<List<Integer>> combinationSum(int[] candidates, int target) {
        List<List<Integer>> results = new ArrayList<>();
        List<Integer> result = new ArrayList<>();
        Set<List<Integer>> hash = new HashSet<>();
        if (candidates == null || candidates.length == 0) {
            return results;
        }
        Arrays.sort(candidates);
        dfsHelper(candidates, 0, target, 0, results,result, hash);
        return results;
    }
    //使用hashset去重
    private void dfsHelper(int[] candidates,
                           int sum,
                           int target,
                           int index,
                           List<List<Integer>> results,
                           List<Integer> result,
                           Set<List<Integer>> hash) {
        if (index > candidates.length ) {
            return;
        }
        if (sum == target) {
            ArrayList<Integer> temp = new ArrayList(result);
            if (!hash.contains(temp)) {
                results.add(temp);
                hash.add(temp);
            }
        }
        for (int i = index; i < candidates.length; i++) {
            //由于是排序数组，如果sum + candidates已经超过target，则不用往下搜索
            if (sum + candidates[i] > target) {
                break;
            }
            sum += candidates[i];
            result.add(candidates[i]);
            dfsHelper(candidates, sum, target, i, results, result, hash);
            sum -= result.get(result.size() - 1);
            result.remove(result.size() - 1);
        }
    } 
}