合并两个排序的链表

Question

I'm trying to merge two sorted linked lists. Here I'm just trying to implement my own algorithm. Of course there are many solutions given on internet. The code is :

Node* MergeLists(Node *headA, Node* headB)
{
    int countA = 0, countB = 0;
    while(headA){countA++; headA = headA->next;}
    while(headB){countB++; headB = headB->next;}
    Node *res, *tres;
    res = new Node();
    res->next = NULL;
    tres = res;
    for(int i = 0; i < countA+countB-1; i++)
    {
        Node* temp = new Node();
        temp->next = NULL;
        tres->next = temp;
        tres = tres->next;
    }
    while(headA != NULL && headB != NULL)
    {
        if(headA->data > headB->data)
        {
            res->data = headB->data;
            res = res->next;
            headB = headB->next;
        }
        else if(headA->data < headB->data)
        {
            res->data = headA->data;
            res = res->next;
            headA = headA->next;
        }
    }
    while(headA)
    {
        res = headA;
    }
    while(headB)
    {
        res = headB;
    }
    return res;
}

This is just a function which returns the head address of merged linked list.

Consider this input / output example :

Input (stdin):

3
4
1 3 5 6
3
2 4 7
1
15
1
12
0
2
1 2

My Output (stdout)

0 0 0 0 0 0 0
0 0
0 0

Expected Output
1 2 3 4 5 6 7
12 15
1 2

So, my output prints all zero. This is due to the problem in this segment of code :

Node *res, *tres;
    res = new Node();
    res->next = NULL;
    tres = res;
    for(int i = 0; i < countA+countB-1; i++)// this is creating a new linked list.
    {
        Node* temp = new Node();
        temp->next = NULL;
        tres->next = temp;
        tres = tres->next;
    }

I think the linkage between the tres and res is not properly happening. Can you please tell me how to correct this problem.

Update :

Node* MergeLists(Node *headA, Node* headB)
{

    int countA = 0, countB = 0;
    Node* tempA, *tempB;
    tempA = headA; tempB = headB;
    while(headA){countA++; tempA = tempA->next;}
    while(headB){countB++; tempB = tempB->next;}
    Node *res, *tres;
    res = new Node();
    res->next = NULL;
    tres = res;
    for(int i = 0; i < countA+countB-1; i++)
    {
        Node* temp = new Node();
        temp->next = NULL;
        tres->next = temp;
        tres = tres->next;
    }
    while(headA != NULL && headB != NULL)
    {
        if(headA->data > headB->data)
        {
            res->data = headB->data;
            res = res->next;
            headB = headB->next;
        }
        else if(headA->data < headB->data)
        {
            res->data = headA->data;
            res = res->next;
            headA = headA->next;
        }
    }
    if(headA)
    {
        res= headA;
        //res = res->next;
        //headA = headA->next;
    }
    if(headB)
    {
        res = headB;
        //res = res->next;
        //headB = headB->next;
    }
    return res;
}

This time ~ no response on stdout ~

Answer 1

Issue is with this part of the code.

while(headA){countA++; headA = headA->next;}
while(headB){countB++; headB = headB->next;}

After this loop is executed, both headA & headB are pointing to NULL; So when

while(headA != NULL && headB != NULL)

loop starts, it doesn't even enter the loop. Since this loop is responsible for assigning values and it doesn't enter this loop. Hence all your values are set to default 0.

As @Slava mentioned you don't even need this loop. As your directly iterating over Nodes and can stop when NULL occurs.

Create a temp pointer and use that pointer to calculate countA and countB.

Something like

Node* temp;
temp = headA;
while(temp){countA++; temp = temp->next;}
temp = headB;
while(temp){countB++; temp = temp->next;}

Also, this may result in infinite loop. Please increment the Node inside the loop. or just change the condition to an if, instead of while.

    while(headA)
    {
        res = headA;
    }
    while(headB)
    {
        res = headB;
    }

Update -

Another Issue: Your returning res after this computation, which is pointing to the last element. Hence the output would be just one digit.

What you can do is

tres = res;
for(int i = 0; i < countA+countB-1; i++)
{
    ...
}
tres = res; // Add this line, so your keeping track of the original head
while(headA != NULL && headB != NULL)

And finally return tres; instead of return res; With this you'll be returning the original head.

Answer 2

You do not need to count elements, precreate result list and use so many loops, do everything in one loop:

Node* MergeLists(Node *headA, Node* headB)
{
    Node *res = nullptr;
    Node **ptr = &res;
    while( headA || headB  ) {
        Node *curr = new Node;
        curr->next = nullptr;
        *ptr = curr;
        ptr = &curr->next;
        if( headB == nullptr || ( headA && headA->data < headB->data ) ) {
            curr->data = headA->data;
            headA = headA->next;
        } else {
            curr->data = headB->data;
            headB = headB->next;
        }
    }
    return res;
}

Note: you should not return raw pointer from a function - it can easily lead to a memory leak. You should use smart pointer instead.

Answer 3

节点应包含某些内容-您正在创建一个新节点并将其分配到链的末尾，但根本不保存任何值，因此该值可能只是默认为零。

Answer 4

        res = res->next; <---

while(headA)
{
    res = headA;
}
while(headB)
{
    res = headB;
}
return res;

you returning res that is pointer to the end instead of the head. Plus while(headB/headA) suppose to add reminder at the end of list but actually will loop infinite because no increment of the pointer.