题解 | #KiKi学结构体和指针#

#include <stdio.h>
#include <stdlib.h>
typedef struct Node {
    int value;
    struct Node* next;
} Node;
Node* buyNewNode() {
    Node* newnode = (Node*)malloc(sizeof(Node));
    newnode->next = NULL;
    scanf("%d ", &(newnode->value));
    return newnode;
}
Node* NodeInit() {
    Node* newnode = (Node*)malloc(sizeof(Node));
    newnode->next = NULL;
    return newnode;
}
//用于输入数据-尾添
void tailPop(Node* head) {
    while (head->next != NULL) {
        head = head->next;
    }
    Node* newnode = buyNewNode();
    head->next = newnode;
}
void del(Node* head, int num, int* last) {
    Node* now = head;
    Node* tmp = NULL;
    *last = 0;
    char flag = 0;//判断head是否找到要删除的数据
    while (head->next != NULL) {
        head = head->next;
        (*last)++;
        while (head->value == num) {
            tmp = head;
            if (head->next != NULL) {
                head = head->next;

            } else if (head->next == NULL) {
                now->next = NULL;
                (*last)--;
                free(head);
                return;
            }
            free(tmp);
            flag = 1;
        }
        if (flag == 0) {
            now = head;
        } else {
            now->next = head;
            now = head;
            flag = 0;
        }

    }
}
void printNode(Node* head) {
    while (head->next != NULL) {
        head = head->next;
        printf("%d ", head->value);
    }
    printf("\n");
}
int main() {
    int n, delnum, last;//last:剩余个数
    Node* head = NodeInit();
    scanf("%d", &n);
    for (int i = 0; i <= n - 1; i++) {
        tailPop(head);
    }
    scanf("%d", &delnum);
    del(head, delnum, &last);
    printf("%d\n", last);
    printNode(head);

}

这道题的难点在于：

当删除最后一个节点时，上一个节点（非要删除的数字）的next应为NULL。

虽然free可以释放你删除的节点，但上一个节点的next仍保留着指针，在用node->next往下走时，会非法访问（因为你的节点已经free）。

所以就要使其next为NULL。