这个链表删除尾节点功能有什么问题?

时间:2010-10-16 10:13:38

标签: c singly-linked-list

我编写了这个函数来删除单链表的最后一个节点。

问题是,它能够删除除第一个/起始节点之外的所有节点。

此代码段中缺少什么?

请回答我的具体问题。

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

struct Node
{
    char CharContent;
    struct Node * NextNodePointer;
};
typedef struct Node Node;

#pragma region Prototypes
Node * CreateNewNode(char ch);
Node * AddNode(Node * start, Node * newNode);
void DeleteTailNode(Node * start);
void PrintAllNodes(const Node * start);
#pragma endregion Comments

main()
{
    Node * start = NULL;
    Node * newNode = NULL;

    start = AddNode(start, CreateNewNode('A'));
    start = AddNode(start, CreateNewNode('B'));
    start = AddNode(start, CreateNewNode('C'));
    start = AddNode(start, CreateNewNode('D'));
    PrintAllNodes(start);

    DeleteTailNode(start);
    PrintAllNodes(start);
    DeleteTailNode(start);
    PrintAllNodes(start);
    DeleteTailNode(start);
    PrintAllNodes(start);
    DeleteTailNode(start);
    PrintAllNodes(start);
    DeleteTailNode(start);
    PrintAllNodes(start);

    getch();
}

#pragma region Node * CreateNewNode(char ch)
Node * CreateNewNode(char ch)
{
    struct Node * newNode = (struct Node *) malloc(sizeof(struct Node *));

    newNode->CharContent = ch;
    newNode->NextNodePointer = NULL;

    return newNode;
}
#pragma endregion Comment

#pragma region UnifiedAddNode()
Node * AddNode(Node * start, Node * newNode)
{
    Node * copyOfStart = start;

    if(start == NULL)
    {
        return newNode;
    }
    else
    {
        while(copyOfStart->NextNodePointer != NULL)
        {
            copyOfStart = copyOfStart->NextNodePointer;
        }

        copyOfStart->NextNodePointer = newNode;

        return start;
    }
}
#pragma endregion Comment


void DeleteTailNode(Node * start)
{
    Node * prev = NULL;
    Node * current = start;

    while(current->NextNodePointer != NULL)
    {
        prev = current;
        current = current->NextNodePointer;
    }

    free (current);

    if (prev != NULL)
    {
        prev->NextNodePointer = NULL;
    }
}


#pragma region PrintAllNodes()
void PrintAllNodes(const Node * start)
{
    struct Node * tempRoot = start;

    while(tempRoot != NULL)
    {
        printf("%c, ", tempRoot->CharContent);

        tempRoot = tempRoot->NextNodePointer;
    }

    printf("\n");
}
#pragma endregion Comment

3 个答案:

答案 0 :(得分:3)

您没有检测到start为NULL的情况,即列表为空。

在将其设置为NULL之前,您不想释放下一个节点吗?

如果起始节点是最后一个节点prev在列表遍历完成时将为NULL,但是当发生这种情况时,您要删除(NULL)start->NextNodePointer,当您想要删除start本身时。

尝试:

void DeleteTailNode(Node *& start)
{
    Node * prev = NULL;
    Node * current = start;

    if (start == NULL)
        return;

    while(current->NextNodePointer != NULL)
    {
        prev = current;
        current = current->NextNodePointer;
    }

    if(current != NULL)
        free(current);

    if(current == start)
        start = NULL;

    if(prev != NULL)
        prev->NextNodePointer = NULL;
}

答案 1 :(得分:2)

内部CreateNewNode()

struct Node * newNode = (struct Node *) malloc(sizeof(struct Node *));  
                                                                  ^
                                                                  |  

                                                                 Ouch!!

将其更改为:struct Node * newNode = (struct Node *) malloc(sizeof(struct Node));

编辑2

测试运行HERE

答案 2 :(得分:1)

您如何理解它是否被删除?正如我所看到的,这里没有任何内容被删除..这是一个100%肯定的内存泄漏...你在DeleteTailNode中使用免费...你只是让分配的内存无法访问..

编辑:循环后调用自由(当前)。并且不需要检查,如果current为NULL,则删除NULL指针是安全的。