假设有两个按元素值递增有序排列的线性表A和B,均以单链表作存储结构,请编写算法将A表和B表归并为一个按元素值递减 有序(即非递增有序,允许表中含有值相同的元素)排列的线性表C,并要求利用原装(即A表和

假设有两个按元素值递增有序排列的线性表A和B,均以单链表作存储结构,请编写算法将A表和B表归并为一个按元素值递增 有序(即非递增有序,允许表中含有值相同的元素)排列的线性表C,并要求利用原装(即A表和B表)的结点空间构造C表。

采用的方法:

尾插法

#include <iostream>
#include <cstdlib>
#include <cstdio>
#include <ctime>
using namespace std;

typedef char ElemType;

typedef struct Node{
	ElemType data;
	struct Node *next;
}Node, *LinkList;

LinkList CreateList()
{
	LinkList L;
	ElemType c;
	L = (LinkList)malloc(sizeof(Node));
	L->next = NULL;
	Node *p , *tail;
	tail = L;
	c = getchar();
	while(c != '#')
	{
		p = (Node *)malloc(sizeof(Node));
		p->data = c;
		tail->next = p;
		tail = p;
		c = getchar();
	}
	tail->next = NULL;
	return L;
}

void ShowList(LinkList L)
{
	Node *p;
	p = L->next;
	while(p != NULL)
	{
		cout << p->data << " ";
		p = p->next;
	}
	cout << endl;
}

LinkList MergeList(LinkList LA, LinkList LB)
{
	LinkList LC;
	Node *pa, *pb, *r;
	pa = LA->next;
	pb = LB->next;
	LC = LA;
	LC->next = NULL;
	r = LC;
	while(pa != NULL && pb != NULL)
	{
		if(pa->data <= pb->data)
		{
			r->next = pa;
			r = pa;
			pa = pa->next;
		}
		else
		{
			r->next = pb;
			r = pb;
			pb = pb->next;
		}
		if(pa)
		{
			r->next = pa;
		}
		else
		{
			r->next = pb;	
		}
	}
	return LC;
}

int main()
{
	LinkList LA;
	LA = CreateList();
	getchar();
	LinkList LB;
	LB = CreateList();

	cout << "LA:" << endl;
	ShowList(LA);

	cout << "LB:" << endl;
	ShowList(LB);

	LinkList LC;
	LC = MergeList(LA, LB);
	cout << "MergeList:" << endl;
	ShowList(LC);

	return 0;
}





假设有两个按元素值递增有序排列的线性表A和B,均以单链表作存储结构,请编写算法将A表和B表归并为一个按元素值递减 有序(即非递增有序,允许表中含有值相同的元素)排列的线性表C,并要求利用原装(即A表和B表)的结点空间构造C表。

采用的方法:

即先进行线性表的逆置,在进行合并

#include <iostream>
#include <cstdlib>
#include <cstdio>
#include <ctime>
using namespace std;

typedef char ElemType;

typedef struct Node{
	ElemType data;
	struct Node *next;
}Node, *LinkList;

LinkList CreateList()
{
	LinkList L;
	ElemType c;
	L = (LinkList)malloc(sizeof(Node));
	L->next = NULL;
	Node *p , *tail;
	tail = L;
	c = getchar();
	while(c != '#')
	{
		p = (Node *)malloc(sizeof(Node));
		p->data = c;
		tail->next = p;
		tail = p;
		c = getchar();
	}
	tail->next = NULL;
	return L;
}

void ShowList(LinkList L)
{
	Node *p;
	p = L->next;
	while(p != NULL)
	{
		cout << p->data << " ";
		p = p->next;
	}
	cout << endl;
}

void ReverseList(LinkList L)
{
	Node *p, *q;
	p = L->next;
	L->next = NULL;
	while(p != NULL)
	{
		q = p->next;
		p->next = L->next;
		L->next = p;
		p = q;
	}

}

LinkList MergeList1(LinkList LA, LinkList LB)
{
	LinkList LC;
	Node *pa, *pb, *r;
	pa = LA->next;
	pb = LB->next;
	LC = LA;
	LC->next = NULL;
	r = LC;
	while(pa != NULL && pb != NULL)
	{
		if(pa->data <= pb->data)
		{
			r->next = pa;
			r = pa;
			pa = pa->next;
		}
		else
		{
			r->next = pb;
			r = pb;
			pb = pb->next;
		}
		if(pa)
		{
			r->next = pa;
		}
		else
		{
			r->next = pb;	
		}
	}
	return LC;
}

LinkList MergeList2(LinkList LA, LinkList LB)
{
	ReverseList(LA);
	ReverseList(LB);
	LinkList LC;
	Node *pa, *pb, *r;
	pa = LA->next;
	pb = LB->next;
	LC = LA;
	LC->next = NULL;
	r = LC;
	while(pa != NULL && pb != NULL)
	{
		if(pa->data <= pb->data)
		{
			r->next = pb;
			r = pb;
			pb = pb->next;
		}
		else
		{
			r->next = pa;
			r = pa;
			pa = pa->next;
		}
		if(pa)
		{
			r->next = pa;
		}
		else
		{
			r->next = pb;	
		}
	}
	return LC;
}

int main()
{
	LinkList LA;
	LA = CreateList();
	getchar();
	LinkList LB;
	LB = CreateList();

	cout << "LA:" << endl;
	ShowList(LA);

	cout << "LB:" << endl;
	ShowList(LB);

	LinkList LC;
	LC = MergeList2(LA, LB);
	cout << "MergeList2:" << endl;
	ShowList(LC);

	return 0;
}



原文地址:https://www.cnblogs.com/lgh1992314/p/5835156.html