Leetcode 39. Combination Sum

Description:

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.
• 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]
]

Thinking:

We use the comination template. But we should notice that the given array is a set, so there is no duplicated number. Then we don't need sort the array and we should note that the return condition is target == 0.

Step:

1. check whether input is valid.
2. create helper function. And decide the return condition.

   Code:
   

   public class Solution {
    public List<List<Integer>> combinationSum(int[] candidates, int target) {
        if(candidates == null || candidates.length == 0) return new ArrayList<List<Integer>>();
        List<List<Integer>> result = new ArrayList<>();
        List<Integer> list = new ArrayList<>();
        combinationSumHelper(result,list,candidates,target, 0);
        return result;
    }
    private void combinationSumHelper(List<List<Integer>> result, List<Integer> list, int[] candidates, int target, int pos){
        if(target == 0){
            result.add(new ArrayList<Integer>(list));
            return;
        }
        if(target < 0){
            return;
        }
        for(int i = pos; i < candidates.length; i++){
            list.add(candidates[i]);
            combinationSumHelper(result,list,candidates,target-candidates[i], i);
            list.remove(list.size()-1);
        }
    }
   }
   

   Conclusion:
   

   we should return whentarget < 0, to avoid stack overflow.