leetcode:39. Combination Sum

Given a set of candidate numbers (C) (without duplicates) 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.
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 {    static List<List<Integer>> ans ;    static int[] candidates;     public List<List<Integer>> combinationSum(int[] candidates, int target) {        ans = new ArrayList<>();        Solution.candidates = candidates;        Arrays.sort(candidates);        ArrayList<Integer> tmp = new ArrayList<>(candidates.length);        dfs(tmp, 0, target);        return ans;    }    public static void dfs(List<Integer> tmp, int curPos, int curVal){        if(curVal == 0){            List<Integer> t = new ArrayList<>(tmp.size() * 2);            for(int i : tmp){                t.add(i);            }ans.add(t);        }        else if(curVal < 0 || curPos >= candidates.length || candidates[curPos] > curVal){            return ;        }        else{            dfs(tmp, curPos+1, curVal);//          //              tmp.add(candidates[curPos]);//              dfs(tmp, curPos + 1, curVal - candidates[curPos]);//              tmp.remove(tmp.size() - 1);            int i = 0;            do{                ++i;                tmp.add(candidates[curPos]);                curVal -= candidates[curPos];                dfs(tmp, curPos + 1, curVal);            }while(curVal > 0);            while(i-- > 0){                tmp.remove(tmp.size() - 1);                curVal += candidates[curPos];            }        }    }}