[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];
}

}
}