Java for LeetCode 040 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, … , a_k) must be in non-descending order. (ie, a₁ ≤ a₂ ≤ … ≤ a_k).
• 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]

解题思路：

修改上题代码，将DFS宽度设置成2即可，注意使用Set，防止重复，JAVA实现如下：

public List<List<Integer>> combinationSum2(int[] candidates, int target) {
		Set<List<Integer>> list = new HashSet<List<Integer>>();
		Arrays.sort(candidates);
		dfs(list, candidates, 0, target, 0);
		return new ArrayList<List<Integer>>(list);
	}
	static List<Integer> list2 = new ArrayList<Integer>();
	static void dfs(Set<List<Integer>> list, int[] array, int result,int target, int depth) {
		if (result == target) {
			list.add(new ArrayList<Integer>(list2));
			return;
		}
		else if (depth >= array.length || result > target)
			return;
		for (int i = 0; i <= 1; i++) {
			for (int j = 0; j < i; j++)
				list2.add(array[depth]);
			dfs(list, array, result + array[depth] * i, target, depth+1);
			for (int j = 0; j < i; j++)
				list2.remove(list2.size() - 1);
		}
	}

 结果453 ms，效率略低，因此换掉Set，用一个变量计算每次DFS的宽度，ＪＡＶＡ实现如下：

public List<List<Integer>> combinationSum2(int[] candidates, int target) {
		ArrayList<List<Integer>> list = new ArrayList<List<Integer>>();
		Arrays.sort(candidates);
		dfs(list, candidates, 0, target, 0);
		return list;
	}
	static List<Integer> list2 = new ArrayList<Integer>();
	static void dfs(ArrayList<List<Integer>> list, int[] array, int result,int target, int depth) {
		if (result == target) {
			list.add(new ArrayList<Integer>(list2));
			return;
		}
		else if (depth >= array.length || result > target)
			return;
		int step=1;
		while(depth<array.length-1&&array[depth]==array[depth+1]){
			depth++;
			step++;
		}	
		for (int i = 0; i <= step; i++) {
			for (int j = 0; j < i; j++)
				list2.add(array[depth]);
			dfs(list, array, result + array[depth] * i, target, depth+1);
			for (int j = 0; j < i; j++)
				list2.remove(list2.size() - 1);
		}
	}