技術標籤：演算法設計

/*
需求示意：
已知有M段不連續的時間段，[t1,t2],[t2,t3],[t3,t4],[t6,t7],[t7,t8],[t9,t10]
合併後應為：[t1,t4]，[t6,t8],[t9,t10],則缺失[t4,t5]的時間段。
條件:時間段前後為閉包，
目的:將滿足條件的零碎片段拼接為一段連續的大片段，並輸出拼接後的片段.
*/

struct  TyTime
{
	int start;
	int end;
};



vector<TyTime> merge(vector<TyTime>& m_vcTyTime)
{
	if (m_vcTyTime.
 
size() <= 1)
	{	 
		return m_vcTyTime;
	}
	vector<TyTime> res;
	//排序，按開始時間從小到大
	sort(m_vcTyTime.begin(), m_vcTyTime.end(), [](TyTime a, TyTime b){return a.start < b.start; });

	int i = 0;
	TyTime temp = { m_vcTyTime[0].start, m_vcTyTime[0].end };
	while (i < m_vcTyTime.size())
	{
		temp.start =
 
 m_vcTyTime[i].start;
		temp.end = m_vcTyTime[i].end;
		while (i < m_vcTyTime.size())
		{
		    //注意陣列越界
			if (i < m_vcTyTime.size() - 1 && temp.end >= m_vcTyTime[i + 1].start)
			{
			 
				if (temp.end < m_vcTyTime[i + 1].end)
				{
				  
					temp.end = m_vcTyTime[i + 1].end;
				}
 


				i++;
			}
			else
			{
				i++;
				break;
			}

		}
		res.push_back(temp);
	}

	return res;
}



void main()
{
	vector<TyTime>t1{ { 1, 2 }, { 2, 3 }, { 3, 4 }, { 5, 6 }, { 6, 7 }, { 8, 9 } };
	cout << "合併前:\n";
	for (auto node : t1)
	{
		cout << "[" << node.start << "," << node.end << "],";
	}
	cout << endl;

	cout << "合併後:\n";
	
	vector<TyTime>t2 = merge(t1);
	for (auto node : t2)
	{
		cout << "[" << node.start << "," << node.end << "],";
	}

	cout << endl;
	system("pause");
}

結果: