本篇内容介绍了“C++怎么实现3个链表排序整合到一起”的有关知识，在实际案例的操作过程中，不少人都会遇到这样的困境，接下来就让小编带领大家学习一下如何处理这些情况吧！希望大家仔细阅读，能够学有所成！
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当两个 递增链表时 可以采用 归并排序 或者 二叉树建立时候的递归。
#include
#include
using namespace std;
typedef struct node
{
int data;
struct node* next;
}NODE;
NODE* CreateList(int a[], int n)
{
NODE* p;
NODE* q;
NODE* head;
p = new NODE;
p->data = a[0];
head = p;
for(int i=1;idata = a[i];
p->next = q;
p = q;
}
p->next = NULL;
return head;
}
//整体思路跟 排序算法中的 归并排序类似
NODE* MerageList(NODE* head_1, NODE* head_2)
{
NODE* head_3 = NULL;
if(head_1 == NULL)
{
head_3 = head_2;
}
else if(head_2 == NULL)
{
head_3 = head_1;
}
else
{
NODE* top1 =head_1;
NODE* top2 =head_2;
NODE* mid = NULL;
if(top1->data < top2->data)
{
head_3 = top1;
mid =head_3 ;
top1 = top1->next;
}
else
{
head_3 = top2;
mid = head_3 ;
top2 = top2->next;
} // 找到合并后头节点
while(top1 && top2)
{
if(top1->data < top2->data)
{
mid->next = top1;
mid = top1;
top1 = top1->next;
}
else
{
mid->next = top2;
mid = top2;
top2 = top2->next;
}
}
if(top1)
{
mid->next = top1;
}
else if(top2)
{
mid->next = top2;
}
}
return head_3;
}
//我们可以采用二叉树插入时的 递归法
NODE* MerageByDG(NODE* head_1, NODE* head_2)
{
if(NULL == head_1)
{
return head_2;
}
else if(NULL == head_2)
{
return head_1;
}
NODE* head_3;
if(head_1->data < head_2->data)
{
head_3 = head_1;
head_3->next = MerageByDG(head_1->next, head_2);
}
else
{
head_3 = head_2;
head_3->next = MerageByDG(head_1, head_2->next);
}
return head_3;
}
int main()
{
int a[]={1,3,5,7};
int c[]={2,4,6,8,10};
NODE* list1 = CreateList(a,4);
NODE* list3 = CreateList(c,5);
list1=MerageByDG(list1, list3);
for(NODE* temp =list1;temp; temp = temp->next)
{
cout<< temp->data<<" ";
}
cout<
普通数据时可用如下方法 用冒泡排序法
#include
#include
using namespace std;
typedef struct node
{
int data;
struct node* next;
}NODE;
NODE* CreateList(int a[], int n)
{
NODE* p;
NODE* q;
NODE* head;
p = new NODE;
p->data = a[0];
head =p;
for(int i=1;idata = a[i];
p->next = q;
p = q;
}
p->next = NULL;
return head;
}
void BubbleSort(NODE* list)
{
for(NODE* temp1= list;temp1; temp1=temp1->next)
{
for(NODE* temp2 =temp1->next; temp2; temp2=temp2->next)
{
if(temp1->data >temp2->data)
{
int temp = temp1->data;
temp1->data = temp2->data;
temp2->data = temp;
}
}
}
}
NODE* Merge (NODE* list1, NODE* list2)
{
NODE* temp2=list2;
NODE* temp1=list1;
NODE* front=NULL;
for(;temp2; temp2 = temp2->next)
{
for(; temp1; front = temp1,temp1=temp1->next)
{
if( ((temp2->data) < (temp1->data)) && (front==NULL))
{
NODE* temp = new NODE;
temp->data = temp2->data;
temp->next = list1;
list1 = temp;
break;
}
else if( (front != NULL) && (front->data < temp2->data) && (temp1->data >= temp2->data) )
{
NODE* temp = new NODE;
temp->data = temp2->data;
temp->next = temp1;
front->next = temp;
break;
}
else if ( ( (temp2->data)>(temp1->data) ) && (NULL==temp1->next) )
{
NODE* temp = new NODE;
temp->data = temp2->data;
temp1->next = temp;
temp->next =NULL;
break;
}
}
temp1 = list1;
front = NULL;
}
return list1;
}
int main()
{
int a[]={5,6,6,7};
int b[]={8,6,4,1};
int c[]={3,5,6,7,8};
NODE* list1 = CreateList(a,4);
NODE* list2 = CreateList(b,4);
NODE* list3 = CreateList(c,5);
list1=Merge(list1, list2);
list1=Merge(list1,list3);
for(NODE* temp =list1;temp; temp = temp->next)
{
cout<< temp->data<<" ";
}
cout<next)
{
cout<< temp->data<<" ";
}
cout<“C++怎么实现3个链表排序整合到一起”的内容就介绍到这里了，感谢大家的阅读。如果想了解更多行业相关的知识可以关注创新互联网站，小编将为大家输出更多高质量的实用文章！
本文标题：C++怎么实现3个链表排序整合到一起
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