LeetCode: Combination Sum 1

LeetCode: Combination Sum 1

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:

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2
[7]
[2, 2, 3]

The source code is as below:

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public class CombinationSum {

    public List<List<Integer>> combinationSum(int[] candidates, int target) {
        // handle special error cases
        if(candidates == null || candidates.length == 0 || target < 0) return null;
        // store the result
        List<List<Integer>> result = new ArrayList<List<Integer>>();
        Arrays.sort(candidates);

        List<Integer> currentList = new ArrayList<Integer>();
        findSum(0, target, currentList, result, candidates);


        return result;
    }

    public void findSum(int currentCursor, int target, List<Integer> currentList, List<List<Integer>> result, int[] candidates) {
        if(target == 0) {
            List<Integer> tmp = new ArrayList<Integer>(currentList);
            result.add(tmp);
            return;
        }

        for(int i = currentCursor; i < candidates.length; i++) {
            int offset = target - candidates[i];
            if(offset < 0) {
                return;
            }
            currentList.add(candidates[i]);
            findSum(i, offset , currentList, result, candidates);
            currentList.remove(currentList.size() - 1);
        }

    }
}