數據結構-C中的鏈接列表庫

Question

我是C的新手，所以我因為卡住而尋求幫助。 創建一個鏈表庫之后，可以從中添加，刪除和打印用戶想要的所有節點，我應該向程序中添加一個整數類型的附加函數，如果該值在其中不存在，則返回-1。鏈接列表。 如果該值確實存在於鏈表中，則它應返回元素的位置。

例如，在此鏈接列表（ a -> b -> c -> d -> NULL ）中，如果我想知道c的位置，它應該返回我一個3，如果我想知道G的位置，它應該返回我我-1

這是我到目前為止可以制作的程序：

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

struct ListNode
{ 
char data;
struct ListNode *nextNode;
};
typedef struct ListNode node;


int main()
   {
    node *startNodePtr= NULL;
    int choice;
    char value;
    printf("\t\t\tLIST OF CHARACTERS\n");
    do
    {
        printf("\n1.Add New Node \t2.Delete Node \t3.Print Current List \t4.QUIT\n\n");//user friendly interface
        scanf("%d",&choice);
        switch(choice)
        {
            case 1: printf("Enter Character: ");
                scanf("\n%c",&value);
                insertNode(&startNodePtr,value);//calling the function to add a node
                break;

            case 2: printf("Delete Character: ");
                scanf("\n%c",&value);
                deleteNode(&startNodePtr,value);//calling the function to remove a node
                break;

            case 3: printList(startNodePtr);//calling the function to list the nodes
                break;

            case 4: continue;//if we type 4 it won't show the default answer

            default:printf("\t\tINVALID ANSWER! Please type 1. 2. 3. or 4.\n");// in case we type any other character that is not 1, 2, 3 or 4. In this way the program will not crash
                break;

        }

    } while(choice!=4);//keep adding or deleting nodes until we enter 4 which refers to "QUIT"

return 0;
 }

void insertNode(node **sPtr, char add)//function to add a node
   {
    node *newPtr;
    node *curPtr;
    node *prevPtr;
    newPtr=malloc(sizeof(node));
    newPtr->data=add;
    newPtr->nextNode=NULL;
    prevPtr=NULL;
    curPtr=*sPtr;

    while(curPtr!=NULL)
    {
        prevPtr=curPtr;
        curPtr=curPtr->nextNode;
    }
    if (prevPtr==NULL)
    {
        *sPtr=newPtr;
    }
    else
    {
        prevPtr->nextNode=newPtr;
    }
 }
void deleteNode(node **sPtr, char remove)//function to remove a node
    {
    node *curPtr;
    node *prevPtr;
    curPtr=*sPtr;
    prevPtr=NULL;

   if(curPtr->data==remove)
   {
        *sPtr=curPtr->nextNode;
        free(curPtr);
        return;
}
while (curPtr!=NULL)
{
    if (curPtr->data==remove)
       {prevPtr->nextNode=curPtr->nextNode;
        free(curPtr);
        return;}
    else
       {prevPtr=curPtr;
       curPtr=curPtr->nextNode;}
}
 }


void printList(node *sPtr)//function to list the nodes
{
if (sPtr==NULL)
{
    printf("The list is empty!\n");
}
else
{
    while (sPtr!=NULL)
    {
        printf("\n%c-->", sPtr->data);
        sPtr=sPtr->nextNode;
    }

}   printf("NULL\n\n");

}

Answer 1

您已經基本解決了它，您只需要計算迭代次數，直到找到正確的元素即可。

int search(struct ListNode *node, char data) {
    int position = 1;
    // List is empty
    if (node == NULL) {
        return -1;
    }
    while (node != NULL) {
        if (node->data == data) {
            return position;
        }
        position++;
        node = node->nextNode;
    }
    // Element was not in the list
    return -1;
}