Combination Sum

题目原型：

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, … , ak) must be in non-descending order. (ie, a1 ≤ a2 ≤ … ≤ ak).
• 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] 

基本思路：

这题比较简单，明显的递归回溯，只要保证下一层的开始元素是这一层的结束元素即可。

	ArrayList<ArrayList<Integer>> list = new ArrayList<ArrayList<Integer>>();

	public ArrayList<ArrayList<Integer>> combinationSum(int[] candidates,
			int target)
	{
		if (candidates == null || candidates.length == 0)
			return list;
		if (target < 0)
			return list;
		//排序数组
		Arrays.sort(candidates);
		combinationSum(candidates, target, new ArrayList<Integer>(), 0);
		return list;
	}

	public void combinationSum(int[] candidates, int target,
			ArrayList<Integer> num, int startIndex)
	{
		if (target < 0)
			return;
		if (target == 0)
		{
			list.add(new ArrayList<Integer>(num));
			return;
		}
		for (int i = startIndex; i < candidates.length; i++)
		{
			num.add(candidates[i]);
			combinationSum(candidates, target - candidates[i], num, i);
			num.remove(num.size() - 1);//必须移除，否则影响下一步
		}
	}

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Combination Sum

原文：http://blog.csdn.net/cow__sky/article/details/22187123