Question

编写一个函数以反向打印双向链表。该功能仅在打印7之后停止，并且不打印列表中的其余项目。我的程序和功能如下。

已编辑以包含未粘贴的代码。对于在腻子复制和粘贴时遇到的问题，我深表歉意。

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

struct node
{
        int data;
        struct node *next;
        struct node *prev;
};
typedef struct node node;


void printRev(node* head);
node* removeNode(node* head, int d);
node* insertFront(node* head, int d);
node* insertBack(node* head, int d);
void print(node* head);
int max(node* head);
int min(node* head);
int locInList(node* head, int x);

int main()
{

        node* head = NULL;

        head = insertFront(head, 5);
        head = insertFront(head, 4);
        head = insertBack(head, 6);
        head = insertBack(head, 7);
        print(head);
        printRev(head);

        printf("Max: %d\n", max(head));
        printf("Min: %d\n", min(head));
        printf("locInList 5: %d\n", locInList(head, 5));
        printf("locInList 9: %d\n", locInList(head, 9));

        head = removeNode(head, 6);
        print(head);
        head = removeNode(head, 4);
        print(head);
        head = removeNode(head, 7);
        print(head);


        return 0;
}

void printRev(node* head) {
        node *cur = head;
        node *tmp = NULL;
        if (cur == NULL) {
            return;
        }
        else {
            while(cur->next != NULL) {
                cur = cur->next;
            }
            while(cur != NULL) {
               printf("%d ", cur->data);
               cur = cur->prev; 
            } 
        }
        printf("\n");
}

node* removeNode(node* head, int d)
{

        node *tmp = head->next;
        head->data = head->next->data;
        head->next = tmp->next;
        free(tmp);
        return head;
}

node* insertFront(node* head, int d)
{
        node *tmp = NULL;
        tmp = malloc(sizeof(node));
        tmp->data = d;
        tmp->next = head;
        head = tmp;
        return head;
}

node* insertBack(node* head, int d)
{
        node *tmp = malloc(sizeof(node));

        tmp->data = d;
        tmp->next = NULL;

        if(head == NULL){
                return head;
        }
  }
        else{
        node *end = head;

        while(end->next != NULL){
                end = end->next;
        }
                end->next = tmp;
            }

        return head;


}

void print(node* head)
{

        node *tmp = head;

        while(tmp != NULL){
   printf("%d ", tmp->data);
                tmp = tmp->next;
                        }
        printf("\n");
}

int max (node* head)
{

        int max = head->data;
        node *tmp = NULL;
        tmp = head;


        while(tmp->next != NULL){
                if(tmp->data >= max){
                        max = tmp->data;
                }
                        tmp = tmp->next;
                }
                     }
        return min;
}

int locInList(node* head, int x)
{

        int i = 0;
        node *tmp = NULL;
        tmp = head;

        while(tmp != NULL){
                if(tmp->data == x){
                return i;
                }else{
                i++;
                tmp = tmp->next;
                        }  }
                return -1;

}

预期结果是-7 6 5 4 收到的结果是-7

Answer 1

insertFront和insertBack都没有设置prev，这是问题的根本原因。（您的反向迭代循环严重取决于正确设置的prev指针。）

Answer 2

由于它是一个双向链接列表，因此您应该将头的后方指针指向函数insertFront中的temp（新插入的一个）。应该是;

 node* insertFront(node* head, int 
d)
{
    node *tmp = NULL;
    tmp = malloc(sizeof(node));
    tmp->data = d;
    tmp->prev=NULL:
     if(head==NULL)
                return tmp;
     head->prev=tmp;
    tmp->next = head;
    return tmp;
}

类似地，在insertBack函数中，请注意使上一个指针指向列表中的上一个节点。

以反向方式打印双向链接列表，仅打印第一个元素

2 个答案: