XMUOJ上机题之链表全部操作

废话不多说，链表全操作全部奉上。

#include <stack> 

void insert(node *(&head), node *(&now), int x){  // 把 x 插入到 head 的末尾。head: 目标链表头指针 now: 目标链表尾指针 x: 插入值
    if(head == NULL){
        head = new node; 
        head->val = x; 
        head->next = NULL; 
        now = head; 
    }
    else{
        node *tmp = new node; 
        tmp->val = x; 
        tmp->next = NULL;
        now->next = tmp; 
        now = tmp; 
    } 
    return ; 
}

node* createList()  // 创建一个链表，返回其头指针
{   node *head = NULL; 
    node *now; 
    int x; 
    while(scanf("%d", &x) && x != -1){
        insert(head, now, x);
    }
    return head; 
}

node* reverseList(node* head) // 返回一个新链表，存储反转后的链表。中间用栈缓存，可以节省内存空间
{
    node *head1 = NULL; 
    node *now; 
    node *p = head; 
    stack <int> stac; 
    while(p != NULL){
        stac.push(p->val); 
        p = p->next; 
    }
    while(!stac.empty()){
        int x = stac.top(); stac.pop();
        insert(head, now, x); 
    }
    return head1; 
}

node* deleteElements(node* head, int v) // 删除所有值为 v 的节点

{ 
    node *p = head; 
    node *pre = head; 
    while(p != NULL){
        if(p->val == v){
            if(p == head){
                head = head->next;
                delete p; 
                p = head; 
                continue; 
            }
            else{
                pre->next = p->next; 
                delete p; 
                p = pre->next; 
                continue; 
            }
        }
        pre = p; 
        p = p->next; 
    }
    return head; 
}

node* merge(node* head1, node* head2) // 合并两个有序链表（从小到大），成为新链表

{//将这个函数补充完整
    node *head = NULL; 
    node *p = head1; 
    node *q = head2; 
    node *now = NULL; 
    while(p != NULL && q != NULL){
        int flag = 1; 
        int x = 0; 
        if(p->val <= q->val){
            x = p->val; 
            flag = 1; 
        }
        else{
            x = q->val; 
            flag = 2; 
        }
        insert(head, now, x);  
        flag == 1 ? p = p->next : q = q->next; 
    }
    while(p != NULL){
        insert(head, now, p->val); 
        p = p->next; 
    }
    while(q != NULL){
        insert(head, now, q->val); 
        q = q->next; 
    }
    return head; 
}