Combination Sum —— LeetCode

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] 

题目大意：给一个候选数组，一个目标值，从候选数组找出所有和等于目标值的List，候选数组元素可以重复。

解题思路：还是用回溯的方式来做，因为可以重复，所以每次都可以从当前元素开始往后依次添加到List，递归判断，然后回溯。

    public List<List<Integer>> combinationSum(int[] candidates, int target) {
        List<List<Integer>> res = new ArrayList<>();
        if (candidates == null || candidates.length == 0) {
            return res;
        }
        List<Integer> tmp = new ArrayList<>();
        helper(res, tmp, 0,candidates, target);
        return res;
    }

    private void helper(List<List<Integer>> res, List<Integer> tmp, int start,int[] candidates, int target) {
        if (0 == target) {
            res.add(new ArrayList<>(tmp));
//            System.out.println(tmp);
            return;
        }
        for (int i = start; i < candidates.length && target >= candidates[i]; i++) {
            tmp.add(candidates[i]);
            helper(res, tmp, i,candidates, target - candidates[i]);
            tmp.remove(tmp.size() - 1);
            int ca = candidates[i];
            while(i<candidates.length&&ca==candidates[i]){
                i++;
            }
        }
    }