C语言反转链表_c语言编写一个程序,反转链表并输出结果

输入：一个链表，以及反转数

输出：反转后的链表，例如，如果链表有9个结点，反转数为3，结果为321654987；反转数为4，结果为432187659；反转数为1，结果不变

函数：

void Reverselist(struct List *plist, int num){
    if(num == 1){
        return;
    }
    int revernum = plist->count/num;
    struct Node *phead = plist->pfront;    //pfront是反转前的头结点
    struct Node *pheadthen;
    struct Node *prearnow;
    int flag = 1;    //flag用来确定反转后链表的头
    while(revernum > 0){
        pheadthen = onereverse(phead, num - 1);
        if(flag == 1){
            plist->pfront = pheadthen;
            flag = 0;
        }else{
            prearnow->pnext = pheadthen;
        }
        prearnow = phead;
        phead = phead->pnext;
        revernum --;
    }
    return;
}

里面的onereverse函数：

struct Node* onereverse(struct Node *phead, int renum){
    struct Node *p1,*p2,*p3;
    p1 = phead;
    p2 = p1->pnext;
    p3 = p2->pnext;
    while(renum > 0){
        p2->pnext = p1;
        p1 = p2;
        p2 = p3;
        if(renum == 1){
            renum = 1;
        }else{
            p3 = p3->pnext;
        }
        renum --;
    }
    phead->pnext = p2;
    return p1;
}