[leedcode 40] Combination Sum II

Given a collection of candidate numbers (C) and a target number (T), find all unique combinations in C where the candidate numbers sums to T.

Each number in C may only be used once in the combination.

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 10,1,2,7,6,1,5 and target 8, 
A solution set is: 
[1, 7] 
[1, 2, 5] 
[2, 6] 
[1, 1, 6] 

public class Solution {
    List<Integer> seq;
    List<List<Integer>> res;
    public List<List<Integer>> combinationSum2(int[] candidates, int target) {
        //和之前的Combination Sum的不同是，这道题每个数字只可以使用一次。因此递归里传的值是i+1（2）
        //因为要进行去重，因此在结果中add时，需要判断结果中是否已经有值了（1）。level代表每次查找时的起始下标
        Arrays.sort(candidates);
        seq=new ArrayList<Integer>();
        res=new ArrayList<List<Integer>>();
        find(candidates,target,0,0);

        return res;
    }
    public void find(int[] candidates,int target,int sum,int level){
        if(sum==target){//1
            if(!res.contains(seq))
            res.add(new ArrayList<Integer>(seq));
            return;
        }
        if(sum>target){
            return;
        }
        for(int i=level;i<candidates.length;i++){
            seq.add(candidates[i]);
            sum+=candidates[i];
            find(candidates,target,sum,i+1);//2
            seq.remove(seq.size()-1);
            sum-=candidates[i];
        }
        
        
        
    }
}