题解 | 牛牛的单向链表

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

// write your code here......
typedef struct array{
    int num;
    struct array *next;
}node;
void print(node *head){
    node*p=head;
    while(p!=NULL){
        printf("%d ",p->num);
        p=p->next;
    }
}
void allclean(node **list){
    node *p=(*list);
    while (p!=NULL) {
    (*list)->next=p->next;
    free(p);
    p=(*list)->next;
    }
    free(*list);
    (*list)=NULL;
}
int main() {

    int n;
    scanf("%d",&n);

    int* arr=(int*)malloc(n*sizeof(int));

    for (int i = 0; i < n; i++) {
        scanf("%d",&arr[i]);
    }

    // write your code here......
    node *head,*tail,*p;
    head=tail=NULL;
    for(int i=0;i<n;i++){
    p=(node*)malloc(sizeof(node));
    p->num=arr[i];
    p->next=NULL;
    if(head==NULL){
        head=p;
    }
    else {
    tail->next=p;
    }
    tail=p;
    }
    print(head);
    allclean(&head);
    free(arr);
    return 0;
}