Question

更新。它在FOR LOOP中为65,519工作。如果我将它增加到65,520，则失败。完全奇怪。

此程序不适用于大输入。它非常适合小输入。我在Xcode上遇到异常。

Thread 1 : EXC_BAD_ACCESS (code=2, address = 0x7fff5f3fffb8).

请告诉我如何绕过这个奇怪的错误。

#include <stdio.h>
#include <stdlib.h>
#include <iostream>
using namespace std;
typedef struct Node * nodePtr;

struct Node{

    int data;
    nodePtr next;

};

nodePtr globalHead;

void partition(nodePtr head, nodePtr *front, nodePtr *back){

    nodePtr fast;
    nodePtr slow;

    if (head == NULL || head->next == NULL){

        *front = head; // &a
        *back = NULL; // &b

    }else{

        slow = head;
        fast = head->next;

        while(fast != NULL){

            fast = fast->next;

            if(fast != NULL){

                slow = slow->next;
                fast = fast->next;

            }

        }

        *front = head; // a
        *back = slow->next; // b
        slow->next = NULL;
        //printList(*front);
        //printList(*back);

    }

}

nodePtr mergeLists(nodePtr a, nodePtr b){

    nodePtr mergedList = NULL;

    if (a == NULL){
        return b;
    }else if (b == NULL){
        return a;
    }

        try {



    if (a->data <= b->data){
        mergedList = a;
        mergedList->next = mergeLists(a->next, b);
    }else{
        mergedList = b;
        mergedList->next = mergeLists(a, b->next);
    }
    }
    catch (int e) {
        cout << "Error is . . " << e << endl;
    }

    return mergedList;

}

void mergeSort(nodePtr *source){

    nodePtr head = *source;
    nodePtr a = NULL;
    nodePtr b = NULL;

    if(head == NULL || head->next == NULL){

        return;

    }

    partition(head, &a, &b);

    mergeSort(&a);
    mergeSort(&b);

    *source = mergeLists(a, b);

}

void push(nodePtr *head, int data){

    nodePtr newNode = (nodePtr) malloc(sizeof(struct Node));
    newNode->data = data;
    newNode->next = NULL;

    if ((*head) == NULL){
        *head = newNode;
        globalHead = *head;
    }else{
        (*head)->next = newNode;
        *head = newNode;
    }

}

void printList(nodePtr head){

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

}

// *head = head in the main function,
// it is only there to connect the two and
// not let make the function return anything
// passed by reference
// globalHead points to the start of the linked list
// if you are passing the address over here you have to
// make a double pointer over there in the function

int main(void)
{
    nodePtr head = NULL;

    // linked list is formed from top to bottom fashion
    // push is done in constant time O(1)

    long long int i;


    //Pushing 200,000 Elements to the Linked List.
    for(i=1 ; i<=200000 ; i++) {
        push(&head, rand()%200000);
    }



    printList(globalHead);

    mergeSort(&globalHead);

    cout << "After Sorting . . \n";

    printList(globalHead);

    return 0;
}

Answer 1

使用递归mergeLists（）是问题，它将为列表中的每个节点调用自身。尝试更改代码，以便代码循环并将节点附加到最初为空的mergeList，使用指向节点的第二个指针，或者可选地指向指向节点的指针，该指针最初设置为＆amp; mergeList。例如，使用名称pMerge而不是mergeList：

Node * mergeLists(Node *a, Node *b)
{
Node *pMerge = NULL;                    // ptr to merged list
Node **ppMerge = &pMerge;               // ptr to pMerge or prev->next
    if(a == NULL)
        return b;
    if(b == NULL)
        return a;
    while(1){
        if(a->data <= b->data){         // if a <= b
            *ppMerge = a;
            a = *(ppMerge = &(a->next));
            if(a == NULL){
                *ppMerge = b;
                break;
            }
        } else {                        // b <= a
            *ppMerge = b;
            b = *(ppMerge = &(b->next));
            if(b == NULL){
                *ppMerge = a;
                break;
            }
        }
    }
    return pMerge;
}

下面是使用指向列表aList []的指针数组对链表进行排序的快速方法的示例代码，其中aList [i]指向大小为2的幂列表i，使用mergeLists（）。

#define NUMLISTS 32                     // size of aList
Node * mergeSort(NODE *pList)
{
Node * aList[NUMLISTS];                 // array of pointers to lists
Node * pNode;
Node * pNext;
int i;
    if(pList == NULL)                   // check for empty list
        return NULL;
    for(i = 0; i < NUMLISTS; i++)       // zero array
        aList[i] = NULL;
    pNode = pList;                      // merge nodes into array
    while(pNode != NULL){
        pNext = pNode->next;
        pNode->next = NULL;
        for(i = 0; (i < NUMLISTS) && (aList[i] != NULL); i++){
            pNode = mergeLists(aList[i], pNode);
            aList[i] = NULL;
        }
        if(i == NUMLISTS)
            i--;
        aList[i] = pNode;
        pNode = pNext;
    }
    pNode = NULL;                       // merge array into one list
    for(i = 0; i < NUMLISTS; i++)
        pNode = mergeLists(aList[i], pNode);
    return pNode;
}

C ++中的合并排序单一链接列表无法进行大输入

1 个答案: