leetcode_Combination Sum

描述:

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 (a1a2, … , 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] 

思路:

这种题目,一般要用到递归或回溯两种方法,用回溯法试过,代码规模总是越来越庞大,但最终还是没能通过所有的测试用例,^-^!

用递归的话这题目看着要容易理解的多,每递归一次target要变为target=target-candidates[i],并将开始index赋值为i,当target==0时,条件满足,如果target<candidates[i],这轮循环结束,方法出栈,当前的i++,继续循环。

总之,用递归来解决问题,难想到,也不太容易被看懂。

代码:

public List<List<Integer>> combinationSum(int[] candidates, int target)
	{
		List<List<Integer>>listResult=new ArrayList<List<Integer>>();
		ArrayList<Integer>list=new ArrayList<Integer>();
		Arrays.sort(candidates);
		combination(listResult, list, candidates, target,0);
		return listResult;
		
	}
	public void combination(List<List<Integer>>listResult,
			ArrayList<Integer>list,int[] candidates, int target,int begin)
	{
		if(target==0)
		{
			listResult.add(list);
			return;
		}
		for(int i=begin;i<candidates.length&&target>=candidates[i];i++)
		{
			ArrayList<Integer>copy=new ArrayList<Integer>(list);
			copy.add(candidates[i]);
			combination(listResult, copy, candidates,target- candidates[i],i);
		}
	}


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