Java for LeetCode 039 Combination Sum

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

The same repeated number may be chosen from C unlimited number of times.

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 2,3,6,7 and target 7,
A solution set is:
[7]
[2, 2, 3]

解题思路：

首先说明下题目Bug，实际测试中C中元素是不会重复的，因此降低了不少难度，最简单的实现方法，DFS算法，JAVA实现如下：

	static public List<List<Integer>> combinationSum(int[] candidates,int target) {
		List<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(List<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 <= target / array[depth]; 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);
		}
	}