题解 | #牛牛的单向链表#

#include <stdio.h>
#include <stdlib.h>
typedef struct ListNode {
    int val;
    struct ListNode* next;
} ListNode;

struct ListNode* createNode(int v) {
    struct ListNode* node = malloc(sizeof(struct ListNode));
    node->val = v;
    node->next = NULL;
    return node;
}

int main() {
    int n;
    while (scanf("%d", &n) != EOF) {
        int v;
        scanf("%d", &v);
        struct ListNode* head = createNode(v);
        struct ListNode* tmp = head;
        for(int i = 0; i < n - 1; i++) {
            scanf("%d", &v);
            struct ListNode* node = createNode(v);
            tmp->next = node;
            tmp = tmp->next;
        }
        tmp = head;
        while(tmp != NULL) {
            printf("%d ", tmp->val);
            tmp = tmp->next;
        }
        while(head) {
            tmp = head;
            head = head->next;
            free(tmp);
        }
        free(head);
    }
    return 0;
}