转自 http://www.geeksforgeeks.org/archives/3622
前两种方法时间和空间上最好， 但个人觉得第三种递归的方法比较好理解
Write a SortedMerge() function that takes two lists, each of which is sorted in increasing order, and merges the two together into one list which is in increasing order. SortedMerge() should return the new list. The new list should
be made by splicing
together the nodes of the first two lists.
For example if the first linked list a is 5->10->15 and the other linked list b is 2->3->20, then SortedMerge() should return a pointer to the head node of the merged list 2->3->5->10->15->20.
There are many cases to deal with: either ‘a’ or ‘b’ may be empty, during processing either ‘a’ or ‘b’ may run out first, and finally there’s the problem of starting the result list empty, and building it up while going through ‘a’ and ‘b’.
Method 1 (Using Dummy Nodes)
The strategy here uses a temporary dummy node as the start of the result list. The pointer Tail always points to the last node in the result list, so appending new nodes is easy.
The dummy node gives tail something to point to initially when the result list is empty. This dummy node is efficient, since it is only temporary, and it is allocated in the stack. The loop proceeds, removing one node from either ‘a’ or ‘b’, and adding it to
tail. When
we are done, the result is in dummy.next.
/*Program to alternatively split a linked list into two halves */
#include<stdio.h>
#include<stdlib.h>
#include<assert.h>
/* Link list node */
structnode
{
intdata;
structnode* next;
};
/* Takes two lists sorted in increasing order, and splices their nodes together to make one big sorted list which is returned. */
structnode* SortedMerge(structnode* a, structnode* b)
{
/* a dummy first node to hang the result on */
//dummy 节点作为入口，指向第一个节点
structnode dummy;
/* tail points to the last result node */
structnode* tail = &dummy;
/* so tail->next is the place to add new nodes
to the result. */
dummy.next = NULL;
while(1)
{
if(a == NULL)
{
/* if either list runs out, use the other list */
tail->next = b;
break;
}
elseif (b == NULL)
{
tail->next = a;
break;
}
if(a->data <= b->data)
{
MoveNode(&(tail->next), &a);
}
else
{
MoveNode(&(tail->next), &b);
}
tail = tail->next;
}
return(dummy.next);
}
//这个函数是关键，把源单链表的头结点切下，加入到新单链表的开头
/* UTILITY FUNCTIONS */
/*MoveNode() function takes 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}
Affter calling MoveNode():
source == {2, 3}
dest == {1, 1, 2, 3}
*/
voidMoveNode(structnode** destRef, structnode** sourceRef)
{
/* the front source node */
//newNode 指向第一个节点,*sourceRef的值即为源单链表的第一个节点的地址
structnode* 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;
}
/* Function to insert a node at the beginging of the linked list */
voidpush(struct node** head_ref, int new_data)
{
/* allocate node */
structnode* new_node =
(structnode*) malloc(sizeof(structnode));
/* 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 */
voidprintList(structnode *node)
{
while(node!=NULL)
{
printf("%d ", node->data);
node = node->next;
}
}
/* Drier program to test above functions*/
intmain()
{
/* Start with the empty list */
structnode* res = NULL;
structnode* a = NULL;
structnode* b = NULL;
/* Let us create two sorted linked lists to test the functions
Created lists shall be a: 5->10->15, b: 2->3->20 */
push(&a, 15);
push(&a, 10);
push(&a, 5);
push(&b, 20);
push(&b, 3);
push(&b, 2);
/* Remove duplicates from linked list */
res = SortedMerge(a, b);
printf("\n Merged Linked List is: \n");
printList(res);
getchar();
return0;
}
Method 2 (Using Local References)
This solution is structurally very similar to the above, but it avoids using a dummy node. Instead, it maintains a struct node** pointer, lastPtrRef, that always points to the last pointer of the result list. This solves the same case that the dummy node did
— dealing with the result list when it is empty. If you are trying to build up a list at its tail, either the dummy node or the struct node** “reference” strategy can be used (see Section 1 for details).
structnode* SortedMerge(structnode* a, structnode* b)
{
structnode* result = NULL;
/* point to the last result pointer */
structnode** lastPtrRef = &result;
while(1)
{
if(a == NULL)
{
*lastPtrRef = b;
break;
}
elseif (b==NULL)
{
*lastPtrRef = a;
break;
}
if(a->data <= b->data)
{
MoveNode(lastPtrRef, &a);
}
else
{
MoveNode(lastPtrRef, &b);
}
/* tricky: advance to point to the next ".next" field */
lastPtrRef = &((*lastPtrRef)->next);
}
return(result);
}
|
Method 3 (Using Recursion)
Merge is one of those nice recursive problems where the recursive solution code is much cleaner than the iterative code. You probably wouldn’t want to use the recursive version for production code however, because it will use stack space which is proportional
to the length of the lists.
structnode* SortedMerge(structnode* a, structnode* b)
{
structnode* result = NULL;
/* Base cases */
if(a == NULL)
return(b);
elseif (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);
}