编写一个函数AlternatingSplit（），它接受一个列表，并将其节点分割成两个较小的列表“a”和“b”。子列表应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，另一个子列表应该是1->1->1。
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方法1（简单） 最简单的方法是迭代源列表，从源中提取节点，然后交替地将它们放在“a”和“b”的前面（或开头）。唯一奇怪的是，节点的顺序与源列表中的顺序相反。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 <bits/stdc++.h> 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<<node->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<< "Resultant Linked List 'a' : " ; printList(a); cout<< "Resultant 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<stdio.h> #include<stdlib.h> #include<assert.h> /* 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 ( " Original linked List: " ); printList(head); /* Remove duplicates from linked list */ AlternatingSplit(head, &a, &b); printf ( " Resultant Linked List 'a' " ); printList(a); printf ( " 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
时间复杂性： O（n），其中n是给定链表中的节点数。 方法2（使用虚拟节点） 下面是另一种方法，它按照与源列表相同的顺序构建子列表。代码在生成“a”和“b”列表时使用临时虚拟头节点。每个子列表都有一个“尾部”指针，指向当前的最后一个节点，这样就可以很容易地将新节点附加到每个列表的末尾。虚拟节点为尾部指针提供了最初指向的对象。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”（总是指向列表中的最后一个指针，而不是最后一个节点）来避免伪节点。
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
<script> 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 </script> |
时间复杂度：O（n），其中n是给定链表中的节点数。 资料来源： http://cslibrary.stanford.edu/105/LinkedListProblems.pdf 如果您发现上述代码/算法不正确，请写评论，或者找到更好的方法来解决相同的问题。
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