单链表实现反转的3种方法示例代码

#include<stdio.h>#include<malloc.h>typedef struct Node{ int data; struct Node *pnext;} Node,*pnode;pnode CreateNode(){ pnode phead=(pnode)malloc(sizeof(Node)); if(phead==NULL) {  printf("fail to allocate memory");  return -1; } phead->pnext=NULL; int n; pnode ph=phead; for(int i=0; i<5; i++) {  pnode p=(pnode)malloc(sizeof(Node));  if(p==NULL)  {   printf("fail to allocate memory");   return -1;  }  p->data=(i+2)*19;  phead->pnext=p;  p->pnext=NULL;  phead=phead->pnext; } return ph;}int list(pnode head){ int count=0; printf("遍历结果：/n"); while(head->pnext!=NULL) {  printf("%d/t",head->pnext->data);  head=head->pnext;  count++; } printf("链表长度为：%d/n",count); return count;}pnode reverse2(pnode head)//两两节点之间不断交换{ if(head == NULL || head->next == NULL) return head; pnode pre = NULL; pnode next = NULL; while(head != NULL){  next = head->next;  head->next = pre;  pre = head;  head = next;} return pre;}void reverse1(pnode head,int count)//把链表的节点值放在数组中，倒置数组{ int a[5]= {0}; for(int i=0; i<count,head->pnext!=NULL; i++) {  a[i]=head->pnext->data;  head=head->pnext; } for(int j=0,i=count-1; j<count; j++,i--)  printf("%d/t",a[i]);}pnode reverse3(pnode pre,pnode cur,pnode t)//递归实现链表倒置{ cur -> pnext = pre; if(t == NULL)  return cur; //返回无头节点的指针，遍历的时候注意 reverse3(cur,t,t->pnext);}pnode new_reverse3(pnode head){ //新的递归转置 if(head == NULL || head->next == NULL)  return head; pnode new_node = new_reverse3(head->next); head->next->next = head; head->next = NULL; return new_node; //返回新链表头指针}int main(){ pnode p=CreateNode(); pnode p3=CreateNode(); int n=list(p); printf("1反转之后：/n"); reverse1(p,n); printf("/n"); printf("2反转之后：/n"); pnode p1=reverse2(p); list(p1); p3 -> pnext = reverse3(NULL,p3 -> pnext,p3->pnext->pnext); printf("3反转之后：/n"); list(p3); free(p); free(p1); free(p3); return 0;}