LeetCode_39 Combination Sum

Original problem: 这里写链接内容
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]

Related problem: 17 Letter Combination of a Phone Number 这里写链接内容
40 Combination Sum II 这里写链接内容
77 Combinations 这里写链接内容
216 Combination Sum III 这里写链接内容
254 Factor Combinations

Since this is very basic backtracking problem, we do not put too much analysis in this problem. Just remember sort the original array first and if current element equals to the previous one, use continue to go to next element.

Here is the solution code:

public class Solution {    public List<List<Integer>> combinationSum(int[] candidates, int target) {        List<List<Integer>> res = new ArrayList<List<Integer>>();        if(candidates == null || candidates.length == 0) return res;        List<Integer> cur = new ArrayList<Integer>();        Arrays.sort(candidates);        helper(res, cur, candidates, 0, target);        return res;    }    private void helper(List<List<Integer>> res, List<Integer> cur, int[] candidates, int start, int target){        if(target == 0){            res.add(new ArrayList<Integer>(cur));            return;        }        for(int ii = start; ii < candidates.length; ii++){            if(ii > 0 && candidates[ii] == candidates[ii-1]) continue;            if(candidates[ii] <= target){                cur.add(candidates[ii]);                helper(res, cur, candidates, ii, target-candidates[ii]);                cur.remove(cur.size()-1);            }            else break;        }    }}