leetcode--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] 

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dfs method

public class Solution {
    public List<List<Integer>> combinationSum2(int[] num, int target) {
        List<List<Integer> > combs = new ArrayList<List<Integer>>();
		Arrays.sort(num);
		int len = num.length;
		List<Integer> oneSolution = new ArrayList<Integer>();
		if(len > 0 && target > 0)
			dfs(combs, oneSolution, 0, num, 0, target);
		return combs;
	}
	private void dfs(List<List<Integer> > combs, List<Integer> oneSolution, int start, int[] num, int sum, int target){
		if(sum > target) return; // can not be a solution
		if(sum == target){
			List<Integer> temp = new ArrayList<Integer>();
			temp.addAll(oneSolution);
			combs.add(temp);
			return; //can not do dfs further for this case
		}
		
		for(int i = start; i < num.length; ++i) {
			sum += num[i];
			oneSolution.add(num[i]);
			dfs(combs, oneSolution, i + 1, num, sum, target); //once it finished, we do next dfs
			sum -= num[i];
			oneSolution.remove(oneSolution.size() - 1);
			//since we have finished the dfs starting with index num[start]
			//if there are same number in num, we need to skip dfs for them
			while(i < num.length - 1 && num[i] == num[i + 1])
				++i; 
			//i stops at the last element such that num[i] == num[start]
		}
    }
}

　　

another code

public class Solution {
    public ArrayList<ArrayList<Integer>> combinationSum2(int[] num, int target) {
        List<List<Integer> > result = new ArrayList<List<Integer> >();
        Arrays.sort(num);
        int len = num.length;
        if(len > 0 && num[0] <= target){
            ArrayList<ArrayListWithSum> alist = new ArrayList<ArrayListWithSum>();
            alist.add(new ArrayListWithSum(new ArrayList<Integer>(), 0));
            int index = 0;
            while(index < len && !alist.isEmpty()){
                ArrayList<ArrayListWithSum> temp = new ArrayList<ArrayListWithSum>();
                int theSame = checkSame(num, index);
                while(!alist.isEmpty()){             
                    ArrayListWithSum temp1 = alist.remove(0);
                    ArrayList<Integer> toBeAdded = new ArrayList<Integer>();
                    for(int j = 0; j < theSame - index + 1; ++j){
                        ArrayList<Integer> templist = new ArrayList<Integer>();
                        templist.addAll(temp1.li);
                        templist.addAll(toBeAdded);
                        if(temp1.sum + j * num[index] < target)
                            temp.add(new ArrayListWithSum(templist, temp1.sum + j * num[index]));
                        else if(temp1.sum + j * num[index] == target)
                            result.add(templist);
                        toBeAdded.add(num[index]);
                    }
                }
                alist = temp;
                index = theSame;
            }
        }        
        return result;
    }

    public static int checkSame(int[] arr, int i){
        int len = arr.length;
        int j = i; 
        for(; j < len; ++j){
            if(arr[j] != arr[i])
                break;
        }
        return j;
    }
}

class ArrayListWithSum{
    ArrayList<Integer> li;
    int sum;
    ArrayListWithSum(ArrayList<Integer> li, int sum){
        this.li = li;
        this.sum = sum;
    }
}