c++ 取两个链表的交集
Problem statement: Write a C++ program to find the intersection of two single linked lists.
问题陈述：编写一个C ++程序来查找两个单个链表的交集。
Example:
例：
Let the first linked list be:
6->5->2->9->NULL
Let the second linked list to be:
2->7->NULL
So there intersection is:
2->NULL
Solution
解
Brute force approach:
蛮力法：
One technique can be to traverse all the nodes of a linked list and check whether the traversed node is in the other linked list or not. Such an approach takes O(m*n) times complexity. m, n=length of linked lists.
一种技术可以是遍历链表的所有节点，并检查遍历的节点是否在另一链表中。 这种方法需要O(m * n)乘以复杂度 。 m ， n =链表的长度 。
Efficient approach:
高效的方法：
Efficient approach use to use merge sort.
高效的方法用于使用合并排序 。
Sort both the linked list using merge sort. ( for detailed refer to: Merge sort for single linked lists)
使用合并排序对两个链表进行排序。 (有关详细信息，请参阅： 合并单个链接列表的排序 )
Scan each linked list and build intersection according to following:
扫描每个链表并按照以下步骤建立交集：
IF (list1->data < list2->data) No intersection found Traverse list1 by one step( list1=list1->next) ELSE IF(list1->data ==list2->data) Createnode(list1->data) && append node to the intersection list Traverse list1 by one step( list1=list1->next) Traverse list2 by one step( list2=list2->next) ELSE//list1->data >list2->data No intersection found Traverse list2 by one step( list2=list2->next) END IF-ELSEDisplay the intersection list
显示路口清单
C ++实现查找两个链表的交集 (C++ implementation to find intersection of two linked lists)
#include<bits/stdc++.h>
using namespace std;
class node{
public:
int data; // data field
struct node *next;
};
void display(class node* head){
node* current=head; // current node set to head
//traverse until current node isn't NULL
while(current!=NULL){
printf("%d ",current->data);
current=current->next; // go to next node
}
}
node* creatnode(int d){
node* temp=(node*)malloc(sizeof(node));
temp->data=d;
temp->next=NULL;
return temp;
}
//merging two sorted list
node* mergeList(node* split1,node* split2){
node* newhead=NULL;
//base cases
if(split1==NULL)
return split2;
if(split2==NULL)
return split1;
if(split1->data<=split2->data){
newhead=split1;
newhead->next=mergeList(split1->next,split2);
}
else{
newhead=split2;
newhead->next=mergeList(split1,split2->next);
}
return newhead;
}
//split list for merge sort
void splitList(node* head,node** split1,node** split2){
node* slow=head;
node* fast=head->next;
while(fast!=NULL){
fast=fast->next;
if(fast!=NULL){
slow=slow->next;
fast=fast->next;
}
}
*split1=head;
*split2=slow->next;
//spliting
slow->next=NULL;
}
//merge sort
void mergeSort(node** refToHead){
node* head=*refToHead;
node* split1,*split2;
//base case
if(head==NULL || head->next==NULL){
return;
}
//split the list in two halves
splitList(head,&split1,&split2);
//recursively sort the two halves
mergeSort(&split1);
mergeSort(&split2);
*refToHead=mergeList(split1,split2);
return;
}
node* findIntersection(node* head1, node* head2){
//base case
if(head1==NULL && head2==NULL)
return NULL;
node* head4=NULL,*temp;
//for inserting the first common node
while( head1!=NULL && head2!=NULL && head4==NULL){
if(head1->data<head2->data){
head1=head1->next;
}
//intersection nodes(intersection)
else if(head1->data==head2->data){
head4=creatnode(head1->data);
temp=head4;
head1=head1->next;
head2=head2->next;
}
else{
head2=head2->next;
}
}
//for other common nodes(intersection)
while( head1!=NULL && head2!=NULL){
if(head1->data<head2->data){
head1=head1->next;
}
//intersection nodes
else if(head1->data==head2->data){
temp->next=creatnode(head1->data);
temp=temp->next;
head1=head1->next;
head2=head2->next;
}
else{
head2=head2->next;
}
}
return head4;
}
int main(){
printf("creating the linked list by inserting new nodes at the end\n");
printf("enter 0 to stop building the list, else enter any integer\n");
int k;
node* curr,*temp;
cout<<"enter list1...\n";
scanf("%d",&k);
node* head1=creatnode(k); //buliding list, first node
scanf("%d",&k);
temp=head1;
///inserting at the end//
while(k){
curr=creatnode(k);
temp->next=curr;//appending each node
temp=temp->next;
scanf("%d",&k);
}
cout<<"displaying list1...\n";
display(head1); // displaying the list
cout<<"\nenter list2...\n";
scanf("%d",&k);
node* head2=creatnode(k); //buliding list, first node
scanf("%d",&k);
temp=head2;
///inserting at the end//
while(k){
curr=creatnode(k);
temp->next=curr;//appending each node
temp=temp->next;
scanf("%d",&k);
}
cout<<"displaying list1...\n";
display(head2);
//sort both the lists
mergeSort(&head1);
mergeSort(&head2);
cout<<"\ndisplaying their intersection...\n";
node* head4=findIntersection(head1,head2);
if(head4)
display(head4);
else
cout<<"No intersection found\n";
return 0;
}
Output
输出量
First run:
creating the linked list by inserting new nodes at the end
enter 0 to stop building the list, else enter any integer
enter list1...
6 5 2 9 0
displaying list1...
6 5 2 9
enter list2...
2 7 0
displaying list1...
2 7
displaying their intersection...
2
Second run:
creating the linked list by inserting new nodes at the end
enter 0 to stop building the list, else enter any integer
enter list1...
6 7 8 0
displaying list1...
6 7 8
enter list2...
1 5 2 0
displaying list1...
1 5 2
displaying their intersection...
No intersection found
Example with explanation:
带有说明的示例：
First linked list: 6->5->2->9->NULL
Second linked list: 2->7->NULL
After applying merge sort:
First linked list: 2->5->6->9->NULL
Second linked list: 2->7->NULL
------------------------------------------------------
//for better understanding nodes are represented
//only by respective values
head1=2
head2=2
FindIntersection(2,2):
0th iteration
head1!=NULL && head2!=NULL
head1->data==head2->data //=2
thus head4(head of intersection list) =2
temp=head4=2
intersection list up to now:
2->NULL
head1=2->next=5;
head2=2->next=7;
1st iteration
head1!=NULL && head2!=NULL
head1->data<head2->data //5<7
thushead1=5->next=6;
temp=same as before
intersection list up to now:
2->NULL
head2=same as before;
2nd iteration
head1!=NULL && head2!=NULL
thushead1=6->next=9;
temp=same as before
intersection list up to now:
2->NULL
head2=same as before;
3rd iteration
head1!=NULL && head2!=NULL
head1->data>head2->data //9>7
thushead2=7->next=NULL;
temp=same as before
intersection list up to now:
2->NULL
Head1=same as before;
4th iteration
Head2=NULL
So, iteration stops & intersection list is build:
2->NULL
翻译自: https://www.includehelp.com/cpp-programs/find-intersection-of-two-linked-lists.aspx
c++ 取两个链表的交集
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