有序数组的归并 链表的合并lianbiao

有序数组合并

有两个递增数组，第一个数组的大小为 m+n   有m个有效数，第二个数组大小为n，有效数为n.试将两个数组合并到第一个数组，且保持数组一的有序。

看到这个题特别容易想到链表的合并，但是若按照链表的正序方法，时间复杂度为O(m*n),效率太慢

所以应该从后向前插入。时间复杂度为O(m+n)

如下代码，含注释。

#include
#include
using namespace std;

#define m 4
#define n 5

void ComOrdArray(int* a1, int* a2, int n1,int n2)
{
	assert(a1||a2);//判断是否有效

	size_t i = n1 + n2-1;  //被放值位置
	size_t j = 0;          //n1-j为比较位置

	while (n2&&(j *(a2 + n2 - 1))
			{
				*(a1 + i) = *(a1 + n1 - 1 - j);
				--i;
				++j;
			}
			*(a1 + i) = *(a2 + n2 - 1);   //找到插入位置  插入
			--i;
		}      
		n2--;
	}
	for (size_t num = 0; num < n2;num++)    
	{
		*(a1 + num) = *(a2 + num);
	}
}

int main()
{
	int a[m + n] = {5,6,7,15};
	int aa[n] = {2,3,6,8,10};
	ComOrdArray(a,aa,m,n);
	for (size_t i = 0; i < m + n; ++i)
	{
		cout << a[i] << " ";
	}

	return 0;
}

有序链表合并

两个有序链表的合并在合并时需要注意的不多，但是对于链表的析构必须要处理好。将s2插入到s1中，由于在合并时并没有创建新的节点，所以原先s2的结构会有改变，且s1含有所有的s2节点，所以将s2也指向s1，两个链表实际管理的是同一个，所以在析构时要特别注意。

实现代码：

void _ComSortList( List &s2)
	{
		if (_headlist == NULL)
		{
			_headlist = s2._headlist;
			return;
		}
		Node* cur1 = _headlist;
		Node* cur2 =s2._headlist;
		Node* next2 = cur2;
		Node* prev1 = NULL;
		while (cur2&&cur1)
		{
			if (cur2->_data > cur1->_data)
			{
				prev1 = cur1;
				cur1 = cur1->_next;

			}
			else
			{
				next2 = cur2->_next;
				if (prev1 == NULL)
				{
					_headlist = cur2;
					prev1 = cur2;
				}

				else
					prev1->_next = cur2;
				cur2->_next = cur1;
				cur2 = next2;
			}
		}
		if (cur1 == NULL)
		{
			prev1->_next = cur2;
		}
		s2._headlist = _headlist;
		_count++;
		s2._count--;
	}

具体代码：

#include
#include
using namespace std;

template
class ListNode
{
public:
	ListNode(const T &data)
		:_data(data)
		, _next(NULL)
	{}
	T _data;
	ListNode* _next;
};
template
class List
{
public:
	typedef ListNode Node;
	List()
		:_headlist(NULL)
		,_count(1)
	{}
	void PushBack(const T& data)
	{
		Node* tmp = new Node(data);
		if (_headlist == NULL)
		{
			_headlist = tmp;
		}
		else
		{
			Node* cur = _headlist;
			while (cur->_next)
			{ 
				cur = cur->_next;
			}
			cur->_next = tmp;
		}
	}
	~List()
	{
		if (_headlist&&_count)
		{
			delete _headlist;
		}
		_headlist = NULL;
		_count--;
	}
	void Printf()
	{
		if (_headlist == NULL)
			return;
		Node* cur = _headlist;
		while (cur)
		{
			cout << cur->_data << " ";
			cur = cur->_next;
		}
		cout << endl;
	}
	void RevPrintfList1()//利用stack
	{
		if (_headlist == NULL)
			return;
		Node* cur = _headlist;
		stack st;
		while (cur)
		{
			st.push(cur->_data);
			cur = cur->_next;
		}
		cout << "the relist is :" << endl;
		while (!st.empty())
		{
			cout << st.top() << " ";
			st.pop();
		}
		cout << endl;
	}
	void RevPrintfList2()//递归
	{
		if (_headlist == NULL)
			return;
		cout << "the relist is :" << endl;
		_RevPrintfList2(_headlist);
		cout << endl;

	}

	void _ComSortList( List &s2)
	{
		if (_headlist == NULL)
		{
			_headlist = s2._headlist;
			return;
		}
		Node* cur1 = _headlist;
		Node* cur2 =s2._headlist;
		Node* next2 = cur2;
		Node* prev1 = NULL;
		while (cur2&&cur1)
		{
			if (cur2->_data > cur1->_data)
			{
				prev1 = cur1;
				cur1 = cur1->_next;

			}
			else
			{
				next2 = cur2->_next;
				if (prev1 == NULL)
				{
					_headlist = cur2;
					prev1 = cur2;
				}

				else
					prev1->_next = cur2;
				cur2->_next = cur1;
				cur2 = next2;
			}
		}
		if (cur1 == NULL)
		{
			prev1->_next = cur2;
		}
		s2._headlist = _headlist;
		_count++;
		s2._count--;
	}
private:
	void _RevPrintfList2(Node* cur)
	{
		if (cur == NULL)
			return;
		_RevPrintfList2(cur->_next);
		cout << cur->_data<<" ";

	}
	Node* _headlist;
	int _count;
};

#define n1 4
#define n2 5


void ComSortList()
{
	int a1[n1] = { 5, 6, 7, 15 };
	int a2[n2] = { 2, 3, 6, 8, 10 };
	List s1;
	List s2;
	for (size_t i = 0; i < n1; ++i)
	{
		s1.PushBack(a1[i]);
	}
	s1.Printf();
	for (size_t i = 0; i < n2; ++i)
	{
		s2.PushBack(a2[i]);
	}
	s2.Printf(); 
	cout << endl;

	s1._ComSortList(s2);
	s1.Printf();

}