單鏈表的操作是面試中經常會遇到的問題，今天總結一下反轉的幾種方案：

1 ，兩兩對換
2， 放入陣列，倒置陣列
3， 遞迴實現

程式碼如下：

#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("遍歷結果：\n");
    while(head->pnext!=NULL)
    {
        printf("%d\t",head->pnext->data);
        head=head->pnext;
        count++;
    }
    printf
 
("連結串列長度為：%d\n",count);
    return count;
}
pnode reverse2(pnode head)//兩兩節點之間不斷交換
{
   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)//把連結串列的節點值放在陣列中，倒置陣列
{
    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)//遞迴實現連結串列倒置
{

    cur -> pnext = pre;
    if(t == NULL)
        return cur;   //返回無頭節點的指標，遍歷的時候注意
    reverse3(cur,t,t->pnext);

}

pnode new_reverse3(pnode head){  //新的遞迴轉置

    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;  //返回新連結串列頭指標

}
int main()
{
    pnode p=CreateNode();
    pnode p3=CreateNode();
    int n=list(p);
    printf("1反轉之後：\n");
    reverse1(p,n);
    printf("\n");
    printf("2反轉之後：\n");
    pnode p1=reverse2(p);
    list(p1);
    p3 -> pnext = reverse3(NULL,p3 -> pnext,p3->pnext->pnext);
    printf("3反轉之後：\n");
    list(p3);
    free(p);
    free(p1);
    free(p3);
    return 0;
}

這裡注意： head ->next = pre; 以及 pre = head->next，前者把head->next 指向 pre，而後者是把head->next指向的節點賦值給pre。如果原來head->next 指向 pnext節點，前者則是head重新指向pre，與pnext節點斷開，後者把pnext值賦值給pre，head與pnext並沒有斷開。