Problem
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.
 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 2 3 4 5 6
 [ [1, 7], [1, 2, 5], [2, 6], [1, 1, 6] ]

Leetcode Link
https://leetcode.com/problems/combinationsumii/#/description
Solution
This problem can be solved using DFS:
 Get the result starting with the first number
 Get the result starting with the second number
 …
 Get the result starting with the last number
But we need to sort the array first in order to remove duplicate records.
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28
 public class Solution { public List<List<Integer>> combinationSum2(int[] candidates, int target) { if(candidates == null  candidates.length == 0) { return new ArrayList<>(); } Arrays.sort(candidates); return combinationSum2(candidates, target, new ArrayList<>(), 0); }
private List<List<Integer>> combinationSum2(int[] candidates, int target, List<Integer> prefix, int startPos) { List<List<Integer>> result = new ArrayList<>(); if(target == 0) { result.add(new ArrayList<>(prefix)); } else if(target > 0){ for (int i = startPos; i < candidates.length; i++) { if(i > startPos && candidates[i] == candidates[i  1]) { continue; } prefix.add(candidates[i]); List<List<Integer>> subResult = combinationSum2(candidates, target  candidates[i], prefix, i + 1); prefix.remove(prefix.size()  1);
result.addAll(subResult); } } return result; } }
