Question

我被分配到一个项目的代码，在某种程度上模拟了机场的时间表。这必须使用linkedLists实现。该项应包含Node类和SLL（单链表）类。项目的逻辑是这样的：在主函数中应该有一个每15分钟滴答一次的时钟;我创建了flight.h文件，其中包含了与航班相关的所有信息。让我困惑的只是与mergeSort和quickSort相关的代码的一些部分。该项目要求对所有航班进行分类：1。根据他们的出发时间2.根据他们的出发城市。

#include <stdlib.h>
#include <string>
#include <ctime>

using namespace std;

#ifndef FLIGHT_H_
#define FLIGHT_H_

class Node
{
public:
    Node()
    {
        flightNum = 0;
        gate = 0;
        status = On_time;
        next = NULL;
    }
    enum Flight_status {On_time, Delayed, Departed};
    struct Time {int hour, minutes;}; time;
    string airLine;
    int flightNum;
    string city;
    int gate;
    Flight_status status;
    Node *next;
    friend class SLL;
};

class SLL
{
private:
    Node *head;
    Node *tail;
    int size;
public:
    SLL() {head = tail = NULL; size = 0;}
    ~SLL() {};
    // Member function to add a new Node into a list
    void addNode();
    // Member functions to perform mergeSort over the list
    void split(SLL *, int, int);
    void merge(SLL *, int, int, int);
    // Member functions to perform quickSort over the list
    void partition(SLL *, int, int);
    void swap(Node &, Node &);
    void quickSort(SLL *, int, int);
    //
    void display(SLL *);
};

#endif /* FLIGHT_H_ */

我正在考虑在此声明的基础上构建程序的其余部分。根据这个类的实现，我想要确定的是解决排序问题的方法很好。如果您发现实施此项目有任何问题，请帮助我。提前谢谢。

Answer 1

以下是代码..根据您的问题修改它。

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

/* Link list node */
struct node
{
    int data;
    struct node* next;
};

/* function prototypes */
struct node* SortedMerge(struct node* a, struct node* b);
void FrontBackSplit(struct node* source,
          struct node** frontRef, struct node** backRef);

/* sorts the linked list by changing next pointers (not data) */
void MergeSort(struct node** headRef)
{
  struct node* head = *headRef;
  struct node* a;
  struct node* b;

  /* Base case -- length 0 or 1 */
  if ((head == NULL) || (head->next == NULL))
  {
    return;
  }

  /* Split head into 'a' and 'b' sublists */
  FrontBackSplit(head, &a, &b); 

  /* Recursively sort the sublists */
  MergeSort(&a);
  MergeSort(&b);

  /* answer = merge the two sorted lists together */
  *headRef = SortedMerge(a, b);
}

struct node* SortedMerge(struct node* a, struct node* b)
{
  struct node* result = NULL;

  /* Base cases */
  if (a == NULL)
     return(b);
  else if (b==NULL)
     return(a);

  /* Pick either a or b, and recur */
  if (a->data <= b->data)
  {
     result = a;
     result->next = SortedMerge(a->next, b);
  }
  else
  {
     result = b;
     result->next = SortedMerge(a, b->next);
  }
  return(result);
}

/* UTILITY FUNCTIONS */
/* Split the nodes of the given list into front and back halves,
     and return the two lists using the reference parameters.
     If the length is odd, the extra node should go in the front list.
     Uses the fast/slow pointer strategy.  */
void FrontBackSplit(struct node* source,
          struct node** frontRef, struct node** backRef)
{
  struct node* fast;
  struct node* slow;
  if (source==NULL || source->next==NULL)
  {
    /* length < 2 cases */
    *frontRef = source;
    *backRef = NULL;
  }
  else
  {
    slow = source;
    fast = source->next;

    /* Advance 'fast' two nodes, and advance 'slow' one node */
    while (fast != NULL)
    {
      fast = fast->next;
      if (fast != NULL)
      {
        slow = slow->next;
        fast = fast->next;
      }
    }

    /* 'slow' is before the midpoint in the list, so split it in two
      at that point. */
    *frontRef = source;
    *backRef = slow->next;
    slow->next = NULL;
  }
}

/* Function to print nodes in a given linked list */
void printList(struct node *node)
{
  while(node!=NULL)
  {
   printf("%d ", node->data);
   node = node->next;
  }
}

/* Function to insert a node at the beginging of the linked list */
void push(struct node** head_ref, int new_data)
{
  /* allocate node */
  struct node* new_node =
            (struct node*) malloc(sizeof(struct node));

  /* put in the data  */
  new_node->data  = new_data;

  /* link the old list off the new node */
  new_node->next = (*head_ref);    

  /* move the head to point to the new node */
  (*head_ref)    = new_node;
} 

/* Drier program to test above functions*/
int main()
{
  /* Start with the empty list */
  struct node* res = NULL;
  struct node* a = NULL;

  /* Let us create a unsorted linked lists to test the functions
   Created lists shall be a: 2->3->20->5->10->15 */
  push(&a, 15);
  push(&a, 10);
  push(&a, 5); 
  push(&a, 20);
  push(&a, 3);
  push(&a, 2); 

  /* Sort the above created Linked List */
  MergeSort(&a);

  printf("\n Sorted Linked List is: \n");
  printList(a);           

  getchar();
  return 0;
}

资料来源：http://en.wikipedia.org/wiki/Merge_sort
http://cslibrary.stanford.edu/105/LinkedListProblems.pdf
http://www.geeksforgeeks.org/merge-sort-for-linked-list/

在链接列表上执行合并排序

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