Combination Sum

1/22/2018
The provided array has no duplicate, so no need to remove duplicate. The solution set won't have duplicates as the tree expands.

Solution:

class Solution {
    public List<List<Integer>> combinationSum(int[] candidates, int target) {
        // write your code here
        List<List<Integer>> res=new ArrayList<>();
        if(candidates==null)
        {
            return null;
        }
        //Arrays.sort(candidates);
        List<Integer> subset=new ArrayList<Integer>();
        Arrays.sort(candidates);
        search(subset, 0, res, candidates, target);
        
        return res;
    }
    
    private void search(List<Integer> subset, int startIndex, List<List<Integer>> res, int[] candidates,  int remain)
    {
        if (remain == 0) {
            res.add(new ArrayList<>(subset));
            return;
        }
       
        for (int i = startIndex; i < candidates.length; i++) {
            if (candidates[i] > remain) return;
            subset.add(candidates[i]);
            search(subset, i, res, candidates, remain - candidates[i]);
            subset.remove(subset.size() - 1);
        }
    }
}

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思路：

subsets的马甲。允许重复，所以startIndex是从i开始而不是i+1。

递归返回条件是target == 0, 每一次递归用target - candidates[i]。还有一种情况是target < candidates[i]，需要剔除掉。

本身答案允许重复的数，让candidates里面还有重复的话，那么结果里面肯定会有重复的组合。所以要有如下代码去重复。

  if (i != 0 && candidates[i] == candidates[i - 1]) {  
         continue;  
       }