给定单链表的交替分割 |设置 1

编写一个函数AlternatingSplit()，它接受一个列表并将其节点划分为两个较小的列表“a”和“b”。子列表应该由原始列表中的交替元素组成。因此，如果原始列表是 0->1->0->1->0->1，那么一个子列表应该是 0->0->0，另一个应该是 1->1->1。

方法一（简单）
最简单的方法是遍历源列表并将节点从源中拉出，然后交替地将它们放在 'a' 和 b' 的前面（或开头）。唯一奇怪的部分是节点的顺序与源列表中出现的顺序相反。方法 2 通过跟踪子列表中的最后一个节点在末尾插入节点。

C++

/* C++ Program to alternatively split
a linked list into two halves */
#include 
using namespace std;
 
/* Link list node */
class Node
{
    public:
    int data;
    Node* next;
};
 
/* pull off the front node of
the source and put it in dest */
void MoveNode(Node** destRef, Node** sourceRef) ;
 
/* Given the source list, split its
nodes into two shorter lists. If we number
the elements 0, 1, 2, ... then all the even
elements should go in the first list, and
all the odd elements in the second. The
elements in the new lists may be in any order. */
void AlternatingSplit(Node* source, Node** aRef,
                            Node** bRef)
{
    /* split the nodes of source
    to these 'a' and 'b' lists */
    Node* a = NULL;
    Node* b = NULL;
         
    Node* current = source;
    while (current != NULL)
    {
        MoveNode(&a, ¤t); /* Move a node to list 'a' */
        if (current != NULL)
        {
            MoveNode(&b, ¤t); /* Move a node to list 'b' */
        }
    }
    *aRef = a;
    *bRef = b;
}
 
/* Take the node from the front of
the source, and move it to the front
of the dest. It is an error to call
this with the source list empty.
     
Before calling MoveNode():
source == {1, 2, 3}
dest == {1, 2, 3}
         
After calling MoveNode():
source == {2, 3}    
dest == {1, 1, 2, 3}    
*/
void MoveNode(Node** destRef, Node** sourceRef)
{
    /* the front source node */
    Node* newNode = *sourceRef;
    assert(newNode != NULL);
         
    /* Advance the source pointer */
    *sourceRef = newNode->next;
         
    /* Link the old dest off the new node */
    newNode->next = *destRef;
         
    /* Move dest to point to the new node */
    *destRef = newNode;
}
 
/* UTILITY FUNCTIONS */
/* Function to insert a node at
the beginning of the linked list */
void push(Node** head_ref, int new_data)
{
    /* allocate node */
    Node* new_node = new 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;
}
 
/* Function to print nodes
in a given linked list */
void printList(Node *node)
{
    while(node!=NULL)
    {
    cout<data<<" ";
    node = node->next;
    }
}
 
/* Driver code*/
int main()
{
    /* Start with the empty list */
    Node* head = NULL;
    Node* a = NULL;
    Node* b = NULL;
     
    /* Let us create a sorted linked list to test the functions
    Created linked list will be 0->1->2->3->4->5 */
    push(&head, 5);
    push(&head, 4);
    push(&head, 3);
    push(&head, 2);
    push(&head, 1);                                
    push(&head, 0);
     
    cout<<"Original linked List: ";
    printList(head);
     
    /* Remove duplicates from linked list */
    AlternatingSplit(head, &a, &b);
     
    cout<<"\nResultant Linked List 'a' : ";
    printList(a);        
     
    cout<<"\nResultant Linked List 'b' : ";
    printList(b);        
     
    return 0;
}
 
// This code is contributed by rathbhupendra

C

/*Program to alternatively split a linked list into two halves */
#include
#include
#include
 
/* Link list node */
struct Node
{
    int data;
    struct Node* next;
};
 
/* pull off the front node of the source and put it in dest */
void MoveNode(struct Node** destRef, struct Node** sourceRef) ;
 
/* Given the source list, split its nodes into two shorter lists.
  If we number the elements 0, 1, 2, ... then all the even elements
  should go in the first list, and all the odd elements in the second.
  The elements in the new lists may be in any order. */
void AlternatingSplit(struct Node* source, struct Node** aRef,
                            struct Node** bRef)
{
  /* split the nodes of source to these 'a' and 'b' lists */
  struct Node* a = NULL;
  struct Node* b = NULL;
   
  struct Node* current = source;
  while (current != NULL)
  {
    MoveNode(&a, ¤t); /* Move a node to list 'a' */
    if (current != NULL)
    {
       MoveNode(&b, ¤t); /* Move a node to list 'b' */
    }
  }
  *aRef = a;
  *bRef = b;
}
 
/* Take the node from the front of the source, and move it to the front of the dest.
   It is an error to call this with the source list empty.
    
   Before calling MoveNode():
   source == {1, 2, 3}  
   dest == {1, 2, 3}
       
   After calling MoveNode():
   source == {2, 3}     
   dest == {1, 1, 2, 3}     
*/
void MoveNode(struct Node** destRef, struct Node** sourceRef)
{
  /* the front source node  */
  struct Node* newNode = *sourceRef;
  assert(newNode != NULL);
   
  /* Advance the source pointer */
  *sourceRef = newNode->next;
   
  /* Link the old dest off the new node */
  newNode->next = *destRef;
   
  /* Move dest to point to the new node */
  *destRef = newNode;
}
 
/* UTILITY FUNCTIONS */
/* Function to insert a node at the beginning 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;
}
 
/* Function to print nodes in a given linked list */
void printList(struct Node *node)
{
  while(node!=NULL)
  {
   printf("%d ", node->data);
   node = node->next;
  }
}
 
/* Driver program to test above functions*/
int main()
{
  /* Start with the empty list */
  struct Node* head = NULL;
  struct Node* a = NULL;
  struct Node* b = NULL; 
 
  /* Let us create a sorted linked list to test the functions
   Created linked list will be 0->1->2->3->4->5 */
  push(&head, 5);
  push(&head, 4);
  push(&head, 3);
  push(&head, 2);
  push(&head, 1);                                   
  push(&head, 0); 
 
  printf("\n Original linked List:  ");
  printList(head);
 
  /* Remove duplicates from linked list */
  AlternatingSplit(head, &a, &b);
 
  printf("\n Resultant Linked List 'a' ");
  printList(a);           
 
  printf("\n Resultant Linked List 'b' ");
  printList(b);           
 
  getchar();
  return 0;
}

Python

# Python program to alternatively split
# a linked list into two halves
 
# Node class
class Node:
     
    def __init__(self, data, next = None):
         
        self.data = data
        self.next = None
 
class LinkedList:
     
    def __init__(self):
         
        self.head = None
     
    # Given the source list, split its
    # nodes into two shorter lists. If we number
    # the elements 0, 1, 2, ... then all the even
    # elements should go in the first list, and
    # all the odd elements in the second. The
    # elements in the new lists may be in any order.
    def AlternatingSplit(self, a, b):
         
        first = self.head
        second = first.next
         
        while (first is not None and
              second is not None and
          first.next is not None):
               
              # Move a node to list 'a'
              self.MoveNode(a, first)
               
              # Move a node to list 'b'
              self.MoveNode(b, second)
               
              first = first.next.next
               
              if first is None:
                break
               
              second = first.next
             
    # Pull off the front node of the
    # source and put it in dest
    def MoveNode(self, dest, node):
         
        # Make the new node
        new_node = Node(node.data)
         
        if dest.head is None:
            dest.head = new_node
        else:
             
            # Link the old dest off the new node
            new_node.next = dest.head
             
            # Move dest to point to the new node
            dest.head = new_node
 
    # UTILITY FUNCTIONS
    # Function to insert a node at 
    # the beginning of the linked list
    def push(self, data):
         
        # 1 & 2 allocate the Node &
        # put the data
         
        new_node = Node(data)
         
        # Make the next of new Node as head
        new_node.next = self.head
         
        # Move the head to point to new Node
        self.head = new_node
         
    # Function to print nodes
    # in a given linked list
    def printList(self):
         
        temp = self.head
        while temp:
            print temp.data,
            temp = temp.next
             
        print("")
 
# Driver Code
if __name__ == "__main__":
     
    # Start with empty list
    llist = LinkedList()
    a = LinkedList()
    b = LinkedList()
     
    # Created linked list will be
    # 0->1->2->3->4->5
    llist.push(5)
    llist.push(4)
    llist.push(3)
    llist.push(2)
    llist.push(1)
    llist.push(0)
     
    llist.AlternatingSplit(a, b)
     
    print "Original Linked List: ",
    llist.printList()
     
    print "Resultant Linked List 'a' : ",
    a.printList()
     
    print "Resultant Linked List 'b' : ",
    b.printList()
     
# This code is contributed by kevalshah5

输出：

Original linked List: 0 1 2 3 4 5 
Resultant Linked List 'a' : 4 2 0 
Resultant Linked List 'b' : 5 3 1

时间复杂度： O(n)，其中 n 是给定链表中的节点数。
方法二（使用虚拟节点）
这是一种以与源列表相同的顺序构建子列表的替代方法。该代码在构建 'a' 和 'b' 列表时使用临时虚拟头节点。每个子列表都有一个指向其当前最后一个节点的“尾”指针——这样可以轻松地将新节点附加到每个列表的末尾。虚拟节点给尾指针一些最初指向的东西。在这种情况下，虚拟节点是有效的，因为它们是临时的并在堆栈中分配。或者，可以使用本地“引用指针”（始终指向列表中的最后一个指针而不是最后一个节点）来避免虚拟节点。

C++

void AlternatingSplit(Node* source,
                      Node** aRef, Node** bRef)
{
    Node aDummy;
     
    /* points to the last node in 'a' */
    Node* aTail = &aDummy;
    Node bDummy;
     
    /* points to the last node in 'b' */
    Node* bTail = &bDummy;
    Node* current = source;
    aDummy.next = NULL;
    bDummy.next = NULL;
    while (current != NULL)
    {
        MoveNode(&(aTail->next), ¤t); /* add at 'a' tail */
        aTail = aTail->next; /* advance the 'a' tail */
        if (current != NULL)
        {
            MoveNode(&(bTail->next), ¤t);
            bTail = bTail->next;
        }
    }
    *aRef = aDummy.next;
    *bRef = bDummy.next;
}
 
// This code is contributed
// by rathbhupendra

C

void AlternatingSplit(struct Node* source, struct Node** aRef,
                            struct Node** bRef)
{
  struct Node aDummy;
  struct Node* aTail = &aDummy; /* points to the last node in 'a' */
  struct Node bDummy;
  struct Node* bTail = &bDummy; /* points to the last node in 'b' */
  struct Node* current = source;
  aDummy.next = NULL;
  bDummy.next = NULL;
  while (current != NULL)
  {
    MoveNode(&(aTail->next), ¤t); /* add at 'a' tail */
    aTail = aTail->next; /* advance the 'a' tail */
    if (current != NULL)
    {
      MoveNode(&(bTail->next), ¤t);
      bTail = bTail->next;
    }
  }
  *aRef = aDummy.next;
  *bRef = bDummy.next;
}

Java

static void AlternatingSplit(Node source, Node aRef,
                            Node bRef)
{
  Node aDummy = new Node();
  Node aTail = aDummy; /* points to the last node in 'a' */
  Node bDummy = new Node();
  Node bTail = bDummy; /* points to the last node in 'b' */
  Node current = source;
  aDummy.next = null;
  bDummy.next = null;
  while (current != null)
  {
    MoveNode((aTail.next), current); /* add at 'a' tail */
    aTail = aTail.next; /* advance the 'a' tail */
    if (current != null)
    {
      MoveNode((bTail.next), current);
      bTail = bTail.next;
    }
  }
  aRef = aDummy.next;
  bRef = bDummy.next;
}
 
// This code is contributed by rutvik_56

C＃

static void AlternatingSplit(Node source, Node aRef,
                            Node bRef)
{
  Node aDummy = new Node();
  Node aTail = aDummy; /* points to the last node in 'a' */
  Node bDummy = new Node();
  Node bTail = bDummy; /* points to the last node in 'b' */
  Node current = source;
  aDummy.next = null;
  bDummy.next = null;
  while (current != null)
  {
    MoveNode((aTail.next), current); /* add at 'a' tail */
    aTail = aTail.next; /* advance the 'a' tail */
    if (current != null)
    {
      MoveNode((bTail.next), current);
      bTail = bTail.next;
    }
  }
  aRef = aDummy.next;
  bRef = bDummy.next;
}
 
// This code is contributed by pratham_76

Javascript

function AlternatingSplit( source,  aRef,
                             bRef)
{
  var aDummy = new Node();
  var aTail = aDummy; /* points to the last node in 'a' */
  var bDummy = new Node();
  var bTail = bDummy; /* points to the last node in 'b' */
  var current = source;
  aDummy.next = null;
  bDummy.next = null;
  while (current != null)
  {
    MoveNode((aTail.next), current); /* add at 'a' tail */
    aTail = aTail.next; /* advance the 'a' tail */
    if (current != null)
    {
      MoveNode((bTail.next), current);
      bTail = bTail.next;
    }
  }
  aRef = aDummy.next;
  bRef = bDummy.next;
}
 
// This code contributed by aashish1995

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