leetcode Combination Sum II

Combination Sum II 原题地址：

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] 

排序之后大致思路和Combination Sum一致，只是每个数只能取一次，注意candidate里可能有重复的数字，所以加入boolean数组，防止solution set中有重复。

public class Solution {    private List<List<Integer>> list = new ArrayList<List<Integer>>();public List<List<Integer>> combinationSum2(int[] candidates, int target) {LinkedList<Integer> _list = new LinkedList<Integer>();Arrays.sort(candidates);boolean[] check = new boolean[candidates.length];helpCombine(_list, candidates, check, target, 0);return list;}private void helpCombine(LinkedList<Integer> _list, int[] candidates, boolean[] check, int target, int idx) {if (target == 0) {LinkedList<Integer> temp = (LinkedList<Integer>) _list.clone();list.add(temp);return;}if (idx == candidates.length || candidates[idx] > target) {return;}if (idx == 0 || candidates[idx] != candidates[idx-1] || (candidates[idx] == candidates[idx-1] && check[idx-1])) {_list.add(candidates[idx]);check[idx] = true;helpCombine(_list, candidates, check, target-candidates[idx], idx+1);check[idx] = false;_list.pollLast();}helpCombine(_list, candidates, check, target, idx+1);}}