2019数据结构----单链表真题

思路：

(1)找到中间节点,将原链表一分为二

(2)后半段链表原地逆置

(3)合并链表

#include <stdio.h>
#include <stdlib.h>

//定义节点类型
typedef struct LNode {
    int data;//数据域
    struct LNode *next;//指针域
} LNode, *LinkList;

void tailList(LinkList &l) {
    l = (LinkList) malloc(sizeof(LNode));
    l->next = NULL;
    int x;
    scanf("%d", &x);//s指向新节点，r指向链表尾
    LinkList s, r = l;
    while (x != 9999) {
        s = (LinkList) malloc(sizeof(LNode));//s存储了这个节点的起始地址.s指向此节点
        s->data = x;
        r->next = s;//新节点给尾节点next指针
        r = s;//r指向新的尾部
        scanf("%d", &x);
    }
    r->next = NULL;
}

void printList(LinkList L) {
    L = L->next;
    while (L != NULL) {
        printf("%d", L->data);//打印当前结点数据
        L = L->next;//指向下一个结点
        if (L != NULL) {
            printf(" ");
        }
    }
    printf("\n");
}

void findMiddle(LinkList l, LinkList &l2) {
    l2 = (LinkList) malloc(sizeof(LNode));
    //双指针法遍历,前指针走两次，后指针走一次
    LinkList pcur = l->next;
    LinkList ppre = l->next;
    while (pcur) {
        pcur = pcur->next;
        if (pcur == NULL) {
            break;
        }
        pcur = pcur->next;
        if (pcur == NULL) {
            break;
        }
        ppre = ppre->next;

    }
    l2->next = ppre->next;
    ppre->next = NULL;
}

//三指针逆置链表
void reverseList(LinkList l2) {
    LinkList r, s, t;
    r = l2->next;
    if (r == NULL) {//可能无节点
        return;
    }
    s = r->next;
    if (s == NULL) {
        return;
    }
    t = s->next;
    while (t) {
        //第二个节点指向第一个节点
        s->next = r;
        //然后整体右移
        r = s;
        s = t;
        t = t->next;
    }
    s->next = r;
    l2->next->next = NULL;//逆置后链表尾部为NULL
    l2->next = s;
}

//三指针合并链表
void merge(LinkList l, LinkList l2) {
    LinkList pcur, p, q;
    pcur = l->next;//指向新链表的表尾
    p = pcur->next;
    q = l2->next;//p,q用来遍历两条原链表
    while (p && q) {
        pcur->next = q;
        q = q->next;
        pcur = pcur->next;
        pcur->next = p;
        p = p->next;
        pcur = pcur->next;
    }
    if (p) {
        pcur->next = p;
    }
    if (q) {
        pcur->next = q;
    }
}

int main() {
    LinkList l;
    LinkList l2;//存储第二条单链表头节点
    tailList(l);
    findMiddle(l, l2);
    reverseList(l2);
    merge(l, l2);
    free(l2);
    printList(l);
    return 0;
}