【Leetcode】Combination Sum II

题目链接：https://leetcode.com/problems/combination-sum-ii/

题目：

C) and a target number (T), find all unique combinations in C where the candidate numbers sums to T.

C may only be used once

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.

10,1,2,7,6,1,5 and target 8, 

A solution set is: 

[1, 7] 

[1, 2, 5] 

[2, 6] 

[1, 1, 6] 

思路：

candidate set每个元素只允许选择一次。因为set中可能会有重复的元素导致solution set中可能会有重复解，比如sum=8，candidate set[1,1,7]，就有两个[1,7]解。每个元素只选择一次，可以在Combination Sum 中改为 dspCombination(sum, i+1);，为了解决重复解的问题，我们引入HashSet保存解，最后在将hashSet转为List返回就好了。

算法：

int cl[] = null;
	List<List<Integer>> iLists = new ArrayList<List<Integer>>();
	List<Integer> list = new ArrayList<Integer>();
	HashSet<ArrayList<Integer>> sets = new HashSet<ArrayList<Integer>>();
	int target = 0;

	public List<List<Integer>> combinationSum2(int[] candidates, int target) {
		Arrays.sort(candidates);
		this.cl = candidates;
		this.target = target;
		dspCombination2(0, 0);
		iLists.addAll(sets);  //HashSet 转为 List
		return iLists;
	}

	private void dspCombination2(int sum, int level) {
		if (sum == target) {
			sets.add(new ArrayList<Integer>(list)); //用HashSet保存解
			return;
		} else if (sum > target) {
			return;
		} else {
			for (int i = level; i < cl.length; i++) {
				sum += cl[i];
				list.add(cl[i]);
				dspCombination2(sum, i + 1); //每个元素只允许选择一次
				list.remove(list.size() - 1);
				sum -= cl[i];
			}
		}
	}