[leetcode] 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 set2,3,6,7and target7,
A solution set is:
[7]

[2, 2, 3]

https://oj.leetcode.com/problems/combination-sum/

思路：返回所有可能，只能枚举了，先排序，然后此题每个元素可以取多次，所以递归下一层的时候选取的元素不变（区别Combination Sum II 中递归下一层只能从后面元素选取）。

import java.util.ArrayList;
import java.util.Arrays;

public class Solution {
	public ArrayList<ArrayList<Integer>> combinationSum(int[] candidates,
			int target) {
		ArrayList<ArrayList<Integer>> result = new ArrayList<ArrayList<Integer>>();
		if (candidates == null || candidates.length == 0)
			return result;

		int n = candidates.length;

		Arrays.sort(candidates);
		ArrayList<Integer> list = new ArrayList<Integer>();

		dfs(0, candidates, target, list, result);

		return result;
	}

	private void dfs(int level, int[] a, int num, ArrayList<Integer> list,
			ArrayList<ArrayList<Integer>> result) {
		if (num == 0) {
			result.add(new ArrayList<Integer>(list));
		} else if (num < 0)
			return;
		else {
			for (int i = level; i < a.length; i++) {
				if (a[i] <= num) {
					list.add(a[i]);
					dfs(i, a, num - a[i], list, result);
					list.remove(list.size() - 1);
				}
			}

		}
	}

	public static void main(String[] args) {
		System.out.println(new Solution().combinationSum(
				new int[] { 2, 3, 6, 7 }, 7));
		System.out.println(new Solution().combinationSum(
				new int[] { 1, 3, 6, 7 }, 7));

		System.out.println(new Solution().combinationSum(
				new int[] { 7,3,2 }, 18));
	}

}

参考：

http://blog.csdn.net/linhuanmars/article/details/20828631