10.8 Combination Sum II

Link: https://oj.leetcode.com/problems/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] 

The only difference between 10.8 and 10.7 Combination Sum, is "Each number in C may only be used once in the combination."

Time: O(n!), Space: O(n)

public class Solution {    public ArrayList<ArrayList<Integer>> combinationSum2(int[] num, int target) {        ArrayList<ArrayList<Integer>> result = new ArrayList<ArrayList<Integer>>();Arrays.sort(num);dfs(num, 0, target, new ArrayList<Integer>(), result);return result;    }        public void dfs(int[] num, int start, int target, ArrayList<Integer> tmp, ArrayList<ArrayList<Integer>> result){if(target == 0){    result.add(new ArrayList<Integer>(tmp));}for(int i = start; i < num.length; i++){    if(target < num[i]) return;    if(i>start && num[i] == num[i-1]) continue;    tmp.add(num[i]);    dfs(num, i+1, target-num[i], tmp, result);    tmp.remove(tmp.size()-1);}}}

Note:  

if(i>start && num[i] == num[i-1]) continue;

cannot be coded as 

if(i>0 && num[i] == num[i-1]) continue;

Otherwise, we will have wrong answer:

Input:[1,1], 2

Output:[]

Expected:[[1,1]]