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.

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);        }    } }