合并排序 通常优先用于对链表进行排序。链表的缓慢随机访问性能使得其他一些算法（如quicksort）表现不佳，而其他算法（如heapsort）则完全不可能。
null
让head作为要排序的链表的第一个节点，headRef作为指向head的指针。请注意，我们需要在MergeSort（）中引用head，因为下面的实现更改了下一个链接以对链表进行排序（不是节点上的数据），因此如果原始head上的数据不是链表中的最小值，则必须更改head节点。
MergeSort(headRef)1) If the head is NULL or there is only one element in the Linked List then return.2) Else divide the linked list into two halves. FrontBackSplit(head, &a, &b); /* a and b are two halves */3) Sort the two halves a and b. MergeSort(a); MergeSort(b);4) Merge the sorted a and b (using SortedMerge() discussed here) and update the head pointer using headRef. *headRef = SortedMerge(a, b);
C++
// C++ code for linked list merged sort #include <bits/stdc++.h> using namespace std; /* Link list node */ class Node { public : int data; Node* next; }; /* function prototypes */ Node* SortedMerge(Node* a, Node* b); void FrontBackSplit(Node* source, Node** frontRef, Node** backRef); /* sorts the linked list by changing next pointers (not data) */ void MergeSort(Node** headRef) { Node* head = *headRef; Node* a; 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); } /* See https:// www.geeksforgeeks.org/?p=3622 for details of this function */ Node* SortedMerge(Node* a, Node* b) { 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(Node* source, Node** frontRef, Node** backRef) { Node* fast; Node* slow; 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(Node* node) { while (node != NULL) { cout << node->data << " " ; node = node->next; } } /* 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; } /* Driver program to test above functions*/ int main() { /* Start with the empty list */ Node* res = NULL; 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); cout << "Sorted Linked List is: " ; printList(a); return 0; } // This is code is contributed by rathbhupendra |
C
// C code for linked list merged sort #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); } /* See https:// www.geeksforgeeks.org/?p=3622 for details of this function */ 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; 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 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; } /* Driver 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 ( "Sorted Linked List is: " ); printList(a); getchar (); return 0; } |
JAVA
// Java program to illustrate merge sorted // of linkedList public class linkedList { node head = null ; // node a, b; static class node { int val; node next; public node( int val) { this .val = val; } } node sortedMerge(node a, node b) { node result = null ; /* Base cases */ if (a == null ) return b; if (b == null ) return a; /* Pick either a or b, and recur */ if (a.val <= b.val) { result = a; result.next = sortedMerge(a.next, b); } else { result = b; result.next = sortedMerge(a, b.next); } return result; } node mergeSort(node h) { // Base case : if head is null if (h == null || h.next == null ) { return h; } // get the middle of the list node middle = getMiddle(h); node nextofmiddle = middle.next; // set the next of middle node to null middle.next = null ; // Apply mergeSort on left list node left = mergeSort(h); // Apply mergeSort on right list node right = mergeSort(nextofmiddle); // Merge the left and right lists node sortedlist = sortedMerge(left, right); return sortedlist; } // Utility function to get the middle of the linked list public static node getMiddle(node head) { if (head == null ) return head; node slow = head, fast = head; while (fast.next != null && fast.next.next != null ) { slow = slow.next; fast = fast.next.next; } return slow; } void push( int new_data) { /* allocate node */ node new_node = new node(new_data); /* link the old list off the new node */ new_node.next = head; /* move the head to point to the new node */ head = new_node; } // Utility function to print the linked list void printList(node headref) { while (headref != null ) { System.out.print(headref.val + " " ); headref = headref.next; } } public static void main(String[] args) { linkedList li = new linkedList(); /* * Let us create a unsorted linked list to test the functions * created. The list shall be a: 2->3->20->5->10->15 */ li.push( 15 ); li.push( 10 ); li.push( 5 ); li.push( 20 ); li.push( 3 ); li.push( 2 ); // Apply merge Sort li.head = li.mergeSort(li.head); System.out.print( " Sorted Linked List is: " ); li.printList(li.head); } } // This code is contributed by Rishabh Mahrsee |
Python3
# Python3 program to merge sort of linked list # create Node using class Node. class Node: def __init__( self , data): self .data = data self . next = None class LinkedList: def __init__( self ): self .head = None # push new value to linked list # using append method def append( self , new_value): # Allocate new node new_node = Node(new_value) # if head is None, initialize it to new node if self .head is None : self .head = new_node return curr_node = self .head while curr_node. next is not None : curr_node = curr_node. next # Append the new node at the end # of the linked list curr_node. next = new_node def sortedMerge( self , a, b): result = None # Base cases if a = = None : return b if b = = None : return a # pick either a or b and recur.. if a.data < = b.data: result = a result. next = self .sortedMerge(a. next , b) else : result = b result. next = self .sortedMerge(a, b. next ) return result def mergeSort( self , h): # Base case if head is None if h = = None or h. next = = None : return h # get the middle of the list middle = self .getMiddle(h) nexttomiddle = middle. next # set the next of middle node to None middle. next = None # Apply mergeSort on left list left = self .mergeSort(h) # Apply mergeSort on right list right = self .mergeSort(nexttomiddle) # Merge the left and right lists sortedlist = self .sortedMerge(left, right) return sortedlist # Utility function to get the middle # of the linked list def getMiddle( self , head): if (head = = None ): return head slow = head fast = head while (fast. next ! = None and fast. next . next ! = None ): slow = slow. next fast = fast. next . next return slow # Utility function to print the linked list def printList(head): if head is None : print ( ' ' ) return curr_node = head while curr_node: print (curr_node.data, end = " " ) curr_node = curr_node. next print ( ' ' ) # Driver Code if __name__ = = '__main__' : li = LinkedList() # Let us create a unsorted linked list # to test the functions created. # The list shall be a: 2->3->20->5->10->15 li.append( 15 ) li.append( 10 ) li.append( 5 ) li.append( 20 ) li.append( 3 ) li.append( 2 ) # Apply merge Sort li.head = li.mergeSort(li.head) print ( "Sorted Linked List is:" ) printList(li.head) # This code is contributed by Vikas Chitturi |
C#
// C# program to illustrate merge sorted // of linkedList using System; public class linkedList { node head = null ; // node a, b; public class node { public int val; public node next; public node( int val) { this .val = val; } } node sortedMerge(node a, node b) { node result = null ; /* Base cases */ if (a == null ) return b; if (b == null ) return a; /* Pick either a or b, and recur */ if (a.val <= b.val) { result = a; result.next = sortedMerge(a.next, b); } else { result = b; result.next = sortedMerge(a, b.next); } return result; } node mergeSort(node h) { // Base case : if head is null if (h == null || h.next == null ) { return h; } // get the middle of the list node middle = getMiddle(h); node nextofmiddle = middle.next; // set the next of middle node to null middle.next = null ; // Apply mergeSort on left list node left = mergeSort(h); // Apply mergeSort on right list node right = mergeSort(nextofmiddle); // Merge the left and right lists node sortedlist = sortedMerge(left, right); return sortedlist; } // Utility function to get the // middle of the linked list node getMiddle(node h) { // Base case if (h == null ) return h; node fastptr = h.next; node slowptr = h; // Move fastptr by two and slow ptr by one // Finally slowptr will point to middle node while (fastptr != null ) { fastptr = fastptr.next; if (fastptr != null ) { slowptr = slowptr.next; fastptr = fastptr.next; } } return slowptr; } void push( int new_data) { /* allocate node */ node new_node = new node(new_data); /* link the old list off the new node */ new_node.next = head; /* move the head to point to the new node */ head = new_node; } // Utility function to print the linked list void printList(node headref) { while (headref != null ) { Console.Write(headref.val + " " ); headref = headref.next; } } // Driver code public static void Main(String[] args) { linkedList li = new linkedList(); /* * Let us create a unsorted linked list to test the functions * created. The list shall be a: 2->3->20->5->10->15 */ li.push(15); li.push(10); li.push(5); li.push(20); li.push(3); li.push(2); // Apply merge Sort li.head = li.mergeSort(li.head); Console.Write( " Sorted Linked List is: " ); li.printList(li.head); } } // This code is contributed by Arnab Kundu |
Javascript
<script> // Javascript program to // illustrate merge sorted // of linkedList var head = null ; // node a, b; class node { constructor(val) { this .val = val; this .next = null ; } } function sortedMerge( a, b) { var result = null ; /* Base cases */ if (a == null ) return b; if (b == null ) return a; /* Pick either a or b, and recur */ if (a.val <= b.val) { result = a; result.next = sortedMerge(a.next, b); } else { result = b; result.next = sortedMerge(a, b.next); } return result; } function mergeSort( h) { // Base case : if head is null if (h == null || h.next == null ) { return h; } // get the middle of the list var middle = getMiddle(h); var nextofmiddle = middle.next; // set the next of middle node to null middle.next = null ; // Apply mergeSort on left list var left = mergeSort(h); // Apply mergeSort on right list var right = mergeSort(nextofmiddle); // Merge the left and right lists var sortedlist = sortedMerge(left, right); return sortedlist; } // Utility function to get the middle // of the linked list function getMiddle( head) { if (head == null ) return head; var slow = head, fast = head; while (fast.next != null && fast.next.next != null ) { slow = slow.next; fast = fast.next.next; } return slow; } function push(new_data) { /* allocate node */ var new_node = new node(new_data); /* link the old list off the new node */ new_node.next = head; /* move the head to point to the new node */ head = new_node; } // Utility function to print the linked list function printList( headref) { while (headref != null ) { document.write(headref.val + " " ); headref = headref.next; } } /* Let us create a unsorted linked list to test the functions created. The list shall be a: 2->3->20->5->10->15 */ push(15); push(10); push(5); push(20); push(3); push(2); // Apply merge Sort head = mergeSort(head); document.write( " Sorted Linked List is: " ); printList(head); // This code contributed by umadevi9616 </script> |
输出：
Sorted Linked List is: 2 3 5 10 15 20
时间复杂性： O（n*logn）
空间复杂性： O（n*logn）
方法2： 这种方法更简单，并且使用logn空间。
mergeSort（）：
- 如果链表的大小为1，则返回标题
- 使用乌龟和兔子的方法找到中间点
- 将下一个mid存储在head2中，即右边的子链接列表中。
- 现在让下一个中点为空。
- 对左、右子链表递归调用mergeSort（），并存储左、右链表的新头。
- 根据参数调用merge（），创建左、右子链表的新头，并存储合并后返回的最终头。
- 返回合并linkedlist的最后一个标题。
合并（头1，头2）：
- 使用指针say merged将合并列表存储在其中，并在其中存储虚拟节点。
- 取一个指针temp并为其指定merge。
- 如果head1的数据小于head2的数据，则将head1存储在temp的下一个中，并将head1移动到head1的下一个。
- 否则，将磁头2存储在temp的下一个位置，并将磁头2移动到磁头2的下一个位置。
- 将temp移到temp的下一个位置。
- 重复步骤3、4和5，直到head1不等于null，head2不等于null。
- 现在，将第一个或第二个链表的所有剩余节点添加到合并链表中。
- 返回下一个合并的链接（这将忽略虚拟链接并返回最终合并链接列表的标题）
JAVA
// Java program for the above approach import java.io.*; import java.lang.*; import java.util.*; // Node Class class Node { int data; Node next; Node( int key) { this .data = key; next = null ; } } class GFG { // Function to merge sort static Node mergeSort(Node head) { if (head.next == null ) return head; Node mid = findMid(head); Node head2 = mid.next; mid.next = null ; Node newHead1 = mergeSort(head); Node newHead2 = mergeSort(head2); Node finalHead = merge(newHead1, newHead2); return finalHead; } // Function to merge two linked lists static Node merge(Node head1, Node head2) { Node merged = new Node(- 1 ); Node temp = merged; // While head1 is not null and head2 // is not null while (head1 != null && head2 != null ) { if (head1.data < head2.data) { temp.next = head1; head1 = head1.next; } else { temp.next = head2; head2 = head2.next; } temp = temp.next; } // While head1 is not null while (head1 != null ) { temp.next = head1; head1 = head1.next; temp = temp.next; } // While head2 is not null while (head2 != null ) { temp.next = head2; head2 = head2.next; temp = temp.next; } return merged.next; } // Find mid using The Tortoise and The Hare approach static Node findMid(Node head) { Node slow = head, fast = head.next; while (fast != null && fast.next != null ) { slow = slow.next; fast = fast.next.next; } return slow; } // Function to print list static void printList(Node head) { while (head != null ) { System.out.print(head.data + " " ); head = head.next; } } // Driver Code public static void main(String[] args) { Node head = new Node( 7 ); Node temp = head; temp.next = new Node( 10 ); temp = temp.next; temp.next = new Node( 5 ); temp = temp.next; temp.next = new Node( 20 ); temp = temp.next; temp.next = new Node( 3 ); temp = temp.next; temp.next = new Node( 2 ); temp = temp.next; // Apply merge Sort head = mergeSort(head); System.out.print( "Sorted Linked List is: " ); printList(head); } } |
Python3
# Python program for the above approach # Node Class class Node: def __init__( self ,key): self .data = key self . next = None # Function to merge sort def mergeSort(head): if (head. next = = None ): return head mid = findMid(head) head2 = mid. next mid. next = None newHead1 = mergeSort(head) newHead2 = mergeSort(head2) finalHead = merge(newHead1, newHead2) return finalHead # Function to merge two linked lists def merge(head1,head2): merged = Node( - 1 ) temp = merged # While head1 is not null and head2 # is not null while (head1 ! = None and head2 ! = None ): if (head1.data < head2.data): temp. next = head1 head1 = head1. next else : temp. next = head2 head2 = head2. next temp = temp. next # While head1 is not null while (head1 ! = None ): temp. next = head1 head1 = head1. next temp = temp. next # While head2 is not null while (head2 ! = None ): temp. next = head2 head2 = head2. next temp = temp. next return merged. next # Find mid using The Tortoise and The Hare approach def findMid(head): slow = head fast = head. next while (fast ! = None and fast. next ! = None ): slow = slow. next fast = fast. next . next return slow # Function to print list def printList(head): while (head ! = None ): print (head.data,end = " " ) head = head. next # Driver Code head = Node( 7 ) temp = head temp. next = Node( 10 ); temp = temp. next ; temp. next = Node( 5 ); temp = temp. next ; temp. next = Node( 20 ); temp = temp. next ; temp. next = Node( 3 ); temp = temp. next ; temp. next = Node( 2 ); temp = temp. next ; # Apply merge Sort head = mergeSort(head); print ( "Sorted Linked List is: " ); printList(head); # This code is contributed by avanitrachhadiya2155 |
C#
// C# program for the above approach using System; // Node Class public class Node { public int data; public Node next; public Node( int key) { this .data = key; next = null ; } } class GFG{ // Function to merge sort static Node mergeSort(Node head) { if (head.next == null ) return head; Node mid = findMid(head); Node head2 = mid.next; mid.next = null ; Node newHead1 = mergeSort(head); Node newHead2 = mergeSort(head2); Node finalHead = merge(newHead1, newHead2); return finalHead; } // Function to merge two linked lists static Node merge(Node head1, Node head2) { Node merged = new Node(-1); Node temp = merged; // While head1 is not null and head2 // is not null while (head1 != null && head2 != null ) { if (head1.data < head2.data) { temp.next = head1; head1 = head1.next; } else { temp.next = head2; head2 = head2.next; } temp = temp.next; } // While head1 is not null while (head1 != null ) { temp.next = head1; head1 = head1.next; temp = temp.next; } // While head2 is not null while (head2 != null ) { temp.next = head2; head2 = head2.next; temp = temp.next; } return merged.next; } // Find mid using The Tortoise and The Hare approach static Node findMid(Node head) { Node slow = head, fast = head.next; while (fast != null && fast.next != null ) { slow = slow.next; fast = fast.next.next; } return slow; } // Function to print list static void printList(Node head) { while (head != null ) { Console.Write(head.data + " " ); head = head.next; } } // Driver Code public static void Main(String[] args) { Node head = new Node(7); Node temp = head; temp.next = new Node(10); temp = temp.next; temp.next = new Node(5); temp = temp.next; temp.next = new Node(20); temp = temp.next; temp.next = new Node(3); temp = temp.next; temp.next = new Node(2); temp = temp.next; // Apply merge Sort head = mergeSort(head); Console.Write( "Sorted Linked List is: " ); printList(head); } } // This code is contributed by umadevi9616 |
Javascript
<script> // JavaScript program for the above approach // Node Class class Node { constructor(val) { this .data = val; this .next = null ; } } // Function to merge sort function mergeSort(head) { if (head.next == null ) return head; var mid = findMid(head); var head2 = mid.next; mid.next = null ; var newHead1 = mergeSort(head); var newHead2 = mergeSort(head2); var finalHead = merge(newHead1, newHead2); return finalHead; } // Function to merge two linked lists function merge(head1, head2) { var merged = new Node(-1); var temp = merged; // While head1 is not null and head2 // is not null while (head1 != null && head2 != null ) { if (head1.data < head2.data) { temp.next = head1; head1 = head1.next; } else { temp.next = head2; head2 = head2.next; } temp = temp.next; } // While head1 is not null while (head1 != null ) { temp.next = head1; head1 = head1.next; temp = temp.next; } // While head2 is not null while (head2 != null ) { temp.next = head2; head2 = head2.next; temp = temp.next; } return merged.next; } // Find mid using The Tortoise and The Hare approach function findMid(head) { var slow = head, fast = head.next; while (fast != null && fast.next != null ) { slow = slow.next; fast = fast.next.next; } return slow; } // Function to print list function printList(head) { while (head != null ) { document.write(head.data + " " ); head = head.next; } } // Driver Code var head = new Node(7); var temp = head; temp.next = new Node(10); temp = temp.next; temp.next = new Node(5); temp = temp.next; temp.next = new Node(20); temp = temp.next; temp.next = new Node(3); temp = temp.next; temp.next = new Node(2); temp = temp.next; // Apply merge Sort head = mergeSort(head); document.write( "Sorted Linked List is: <br/>" ); printList(head); // This code contributed by gauravrajput1 </script> |
C++
#include<iostream> using namespace std; //Node structure struct Node{ int data; Node *next; }; //function to insert in list void insert( int x,Node **head) { if (*head == NULL){ *head = new Node; (*head)->data = x; (*head)->next = NULL; return ; } Node *temp = new Node; temp->data = (*head)->data; temp->next = (*head)->next; (*head)->data=x; (*head)->next=temp; } //function to print the list void print(Node *head) { Node *temp=head; while (temp!=NULL) { cout<<temp->data<< " " ; temp = temp->next; } } Node *merge(Node *firstNode,Node *secondNode) { Node *merged = new Node; Node *temp = new Node; //merged is equal to temp so in the end we have the top Node. merged = temp; //while either firstNode or secondNode becomes NULL while (firstNode != NULL && secondNode != NULL) { if (firstNode->data<=secondNode->data) { temp->next = firstNode; firstNode = firstNode->next; } else { temp->next = secondNode; secondNode = secondNode->next; } temp = temp->next; } //any remaining Node in firstNode or secondNode gets inserted in the temp List while (firstNode!=NULL) { temp->next = firstNode; firstNode = firstNode->next; temp = temp->next; } while (secondNode!=NULL) { temp->next = secondNode; secondNode = secondNode->next; temp = temp->next; } //return the head of the sorted list return merged->next; } //function to calculate the middle Element Node *middle(Node *head) { Node *slow = head; Node *fast = head->next; while (slow->next != NULL && (fast!=NULL && fast->next!=NULL)) { slow = slow->next; fast = fast->next->next; } return slow; } //function to sort the given list Node *sort(Node *head){ if (head->next == NULL) { return head; } Node *mid = new Node; Node *head2 = new Node; mid = middle(head); head2 = mid->next; mid->next = NULL; //recursive call to sort() hence diving our problem, and then merging the solution Node *finalhead = merge(sort(head),sort(head2)); return finalhead; } int main( void ) { Node *head = NULL; int n[]={7,10,5,20,3,2}; for ( int i=0;i<6;i++) { insert(n[i],&head); //inserting in the list } cout<< "Sorted Linked List is: " ; print(sort(head)); //printing the sorted list returned by sort() return 0; } |
输出：
Sorted Linked List is: 2 3 5 7 10 20
时间复杂性 ：O（n*logn）
空间复杂性： O（对数n）
资料来源： http://en.wikipedia.org/wiki/Merge_sort 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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