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, … , a_k) must be in non-descending order. (ie, a₁ ≤ a₂ ≤ … ≤ a_k).
• 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 {
    ArrayList<ArrayList<Integer>> result = null;
    public ArrayList<ArrayList<Integer>> combinationSum2(int[] num, int target) {
        // Note: The Solution object is instantiated only once and is reused by each test case.
        Arrays.sort(num);
        result = new ArrayList<ArrayList<Integer>>();
        getSolution(new ArrayList<Integer>(), target, num, 0);
        return result;
    }
    public void getSolution(ArrayList<Integer> row, int target, int[] num, int pos){
        int i = pos;
        int per = -1;
        if(target == 0){
            result.add(row);
        }
        while(i < num.length && num[i] <= target){
            if(per != num[i]){
                per = num[i];
                row.add(num[i]);
                target -= num[i];
                int tmp = num[i];
                getSolution(new ArrayList<Integer>(row), target, num, i + 1);
                row.remove(row.size() - 1);
                target += num[i];
            }
            i ++;
        }
    }
}

 第三遍：

test case:

Input:	[1,1], 1
Output:	[[1],[1]]
Expected:	[[1]]

结果相同如何处理。

public class Solution {
    public List<List<Integer>> combinationSum2(int[] candidates, int target) {
        Arrays.sort(candidates);
        List<List<Integer>> result = new ArrayList<List<Integer>>();
        findElement(candidates, 0, candidates.length, result, new ArrayList<Integer>(), target);
        return result;
    }
    
    public void findElement(int[] arr, int start, int end, List<List<Integer>> result, ArrayList<Integer> row, int tar){
        if(tar == 0) result.add(row);
        for(int i = start; i < end; i ++){
            if((i == start || arr[i] != arr[i - 1]) && arr[i] <= tar){
                ArrayList<Integer> rown = new ArrayList<Integer> (row);
                rown.add(arr[i]);
                findElement(arr, i + 1, end, result, rown, tar - arr[i]);
            }
        }
    }
}

(i == start || arr[i] != arr[i - 1]) 保证了在同一层里不会选取一样的数字。