3.4 从无头单链表中删除节点
假设一个没有头指针的单链表，只有一个指向中间节点（不是头尾节点）的指针，请将该节点从单链表中删除。
思路：假设中间一段链表为 A->B->C，而p指向节点B。因为没有头节点，其实无法追溯回A，然后使A->C。所以这里的技巧是将C中的value值赋给B中，使B成为C，然后删除真正的C，从而达到目的。
1 #include "stdafx.h" 2 #include <iostream> 3 4 using namespace std; 5 6 const int MID_NUM = 3; 7 const int MAX_NUM = 8; 8 9 struct node{ 10 int value; 11 struct node *next; 12 }; 13 14 node* createMidLinkedList(int midNum,node *headPtr){ 15 node *midPtr,*p; 16 17 for(int i=0;i<MAX_NUM;i++){ 18 p = (node *)malloc(sizeof(node)); 19 p->value = i; 20 p->next = headPtr->next; 21 headPtr->next = p; 22 if(p->value==midNum) 23 midPtr = p; 24 cout<<p->value<<" -> "; 25 } 26 cout<<endl<<endl; 27 return midPtr; 28 } 29 30 //delete middle node 31 bool deleteMidNode(node *midPtr){ 32 if(midPtr==NULL || midPtr->next==NULL) return false; 33 34 node *mid,*midNext; 35 mid = midPtr; 36 midNext = mid->next; 37 38 if(midNext!=NULL){ 39 mid->value = midNext->value; 40 mid->next = midNext->next; 41 delete midNext; // 42 } 43 44 45 return true; 46 } 47 48 int _tmain(int argc, _TCHAR* argv[]) 49 { 50 node *head= (node *)malloc(sizeof(node));; //不可在createMidLinkedList方法中分配内存 51 head->next = NULL; //一定不可以漏，否则判断head->next==NULL时会返回false！ 52 node *mid = createMidLinkedList(MID_NUM,head); 53 if(deleteMidNode(mid)){ 54 head = head->next; 55 while(head!=NULL){ 56 cout<<head->value<<" -> "; 57 58 head = head->next; 59 } 60 }else{ 61 cout<<"Failed to delete!"<<endl; 62 } 63 }
输出为：
可见节点3(MID_NUM)已经被删除。
引申问题：只有一个头节点，翻转单链表，要求只遍历一次
利用构造链表时的方法，修改代码如下：
1 node *reverseLinkedList(node *headPtr){ 2 if(headPtr==NULL) return NULL; 3 4 node *head,*newHead,*p,*temp; 5 head=headPtr; 6 p = head->next; 7 newHead = (node *)malloc(sizeof(node)); 8 newHead->next = NULL; 9 10 while(p!=NULL){ 11 temp = p->next; 12 p->next = newHead->next; 13 newHead->next = p; 14 p = temp; 15 } 16 17 return newHead; 18 } 19 20 int _tmain(int argc, _TCHAR* argv[]) 21 { 22 node *head= (node *)malloc(sizeof(node));; //不可在createMidLinkedList方法中分配内存 23 head->next = NULL; //一定不可以漏，否则判断head->next==NULL时会返回false！ 24 node *mid = createMidLinkedList(MID_NUM,head); 25 node *hPtr; //new header pointer 26 if(deleteMidNode(mid)){ 27 hPtr = head; 28 29 //print list 30 head = head->next; 31 while(head!=NULL){ 32 cout<<head->value<<" -> "; 33 head = head->next; 34 } 35 cout<<endl<<endl; 36 37 head = reverseLinkedList(hPtr); 38 39 //print list 40 head = head->next; 41 while(head!=NULL){ 42 cout<<head->value<<" -> "; 43 head = head->next; 44 } 45 46 }else{ 47 cout<<"Failed to delete!"<<endl; 48 } 49 }
可见输出如下：
