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);
}
}