数据结构第讲浙江工业大学详解演示文稿

node() : next(NULL) {}
node(const T& item, node<T> *nextNode = NULL) : nodeValue(item), next(nextNode)
{} };
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1.如何生成链表
node<int> *nodeptr1,*nodeptr2; nodeptr1=new node<int>(1,null); nodeptr2=new node<int>(2,null); 思考：如何连接两个结点？ nodeptr1->next=nodeptr2; nodeptr3=new node<int>(3,null); nodeptr2->next=nodeptr3;
}
else{
newNode=new node<int>(i,NULL);
back->next=newNode;
back=back->next;
}
i++;
}//不同的编程风格
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3. 1 Inserting at the Front of a Linked List
Before
After front
front back item
Before
front
(a)
newNode
20
55
front
After
front
(b)
item
20
55
front
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3.2 Handling the Back of the List
top D
C B A
Disconnect
Reconnect
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Linked List Nodes Removal is like Insertion in reverse.
Disconnect
Reconnect
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Node Composition An individual Node is composed of two parts, a
Data field containing the data stored by the node, and a Pointer field that marks the address of the next Node in the list.
nodeValue
next
nodeValue next
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Abstract Model of a List Object
data next first
data next
data next
data null
0xbfffaa20
0xbfffaa28
0xbfffaa60
0xbfffaa80
2 0xbffBiblioteka Baiduaa28
4 0xbfffaa60
数据结构第讲浙江工业大学详解演示文稿
（优选）数据结构第讲浙江工业大学
Linked List Nodes Each Node is like a piece of a chain
Individual Piece
Pop Chain
To insert a new link, break the chain at the desired location and simply reconnect at both ends of the new piece.
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2. Creating a Linked List
将1,2,3,4,5顺序插入链表（反向插入） node<int> *front=Null, *newNode; int i;
for(i=1;i<=5;i++) {
front=new node<int>(i,front); }
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2. Creating a Linked List
node<int> *front=Null, *newNode,*back;
int i=0;
while(i<5)
{
if(i==0){
front=new node<int>(i,NULL);
back=front;
6 0xbfffaa80
8 Null
first 0xbfffaa20
first是指向node类型的指针： node<int> *first; first=new node<int>(2,NULL);
Node结点设计
Template <typename T> class node {
public: T nodeValue; // data held by the node node<T> *next; // next node in the list
Stack
front
D
C
B
A
Linked List
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3.3 Handling the Back of the List
node<T> *curr=front; while(curr!=null){
curr=curr->next; }
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2. Creating a Linked List
将1,2,3,4,5顺序插入链表（正向插入） node<int> *nodeptr1,*nodeptr2; for(int i=1;i<6;i++) {
nodeptr1=new node<int>(i,null); nodeptr1? } nodeptr1=new node<int>(1,null); for(int i=2;i<6;i++) { nodeptr2=new node<int>(i,null); nodeptr1->next=nodeptr2; nodeptr1=nodeptr1->next; }
3.4 Handling the Back of the List – with back pointer
front back
... //
//
item newNode
The initial list is empty: front=NULL, back=NULL.
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