39. Combination Sum 求目标和的元素集合

39. Combination Sum

问题描述

Given a set of candidate numbers (C) (without duplicates) 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.
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

• 可采用递归和分治的思想
• 和为target的结果 = $\sum_{i=0}^{n} (和为target-candidates[i] 的结果 ,连接, candidates[i])$
• 为了避免结果重复，遍历过程中，要去除已经存在的结果

算法描述

1. 对数组candidates升序排序
2. 调用combinationSumOrder(vector& candidates, int target)递归实现
   1. 若candidates为空，则返回空；
   2. 遍历数组candidates：it
      1. 得到从it开始的新数组 candidates_new，得到新的目标值target_new
      2. 如果target_new <0 ,则返回空
      3. 如果target_new==0，则返回
      4. 如果target_new > 0,则执行combinationSumOrder(candidates_new,target_new),得到上一步的结果result_tmp;
      5. if result_tmp 不为空，则将result_tmp和*it拼接
      6. 再将result_tmp和result整合
3. 得到result。

代码

//39. Combination Sum
vector<vector<int>> Solution::combinationSum(vector<int>& candidates, int target)
{
	vector<vector<int>> result;
	if (candidates.empty())
		return result;
	sort(candidates.begin(), candidates.end());
	result = combinationSumOrder(candidates, target);
	return result;
}
vector<vector<int>> combinationSumOrder(vector<int>& candidates, int target)
{
	// candidates 是有序的
	vector<vector<int>> result;
	if(candidates.empty())
		return result;
	const int n = candidates.size();
	for(auto it =candidates.begin(); it != candidates.end(); it++)
	{
		vector<int> candidates_new(it,candidates.end());
		int target_new=target - *it;
		if(target_new == 0)
		{
			vector<int> tmp;
			result.push_back({*it});
			break;
		}
		else if(target_new < 0)
		{
			break;
		}
		else 
		{
			vector<vector<int>> result_tmp = combinationSumOrder(candidates_new,target_new);
			if(result_tmp.empty())
				continue;
			for (int j = 0; j < result_tmp.size();j++)
			{
				result_tmp[j].insert(result_tmp[j].begin(),*it);
			}
			
			result.insert(result.end(),result_tmp.begin(),result_tmp.end());
			
		}		
	}
	return result;
}