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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;i data = 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;i data = 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!
本文名称:C++怎么实现3个链表排序整合到一起
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