面试题76：单链表的归并排序

#include <iostream>
#include <string>
using namespace std;

/*
MergeSort(headRef)
1) If 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);
*/

struct Node{
int val;
Node *next;
Node(int _val) :val(_val), next(NULL){}
};

void FrontBackSplit(Node *head, Node **a, Node **b)
{
if (head == NULL || head->next == NULL) return;
Node *faster = head->next;
Node *slower = head;
while (faster && faster->next)
{
slower = slower->next;
faster = faster->next->next;
}
*b = slower->next;
slower->next = NULL;
*a = head;
}

Node *SortedMerge(Node *a, Node *b)
{
if (a == NULL) return b;
if (b == NULL) return a;
Node *head = NULL;
if (a->val < b->val)
{
head = a;
a = a->next;
}
else{
head = b;
b = b->next;
}
Node *pNext = head;
while (a && b)
{
if (a->val < b->val)
{
pNext->next = a;
a = a->next;
}
else
{
pNext->next = b;
b = b->next;
}
pNext = pNext->next;
}
if (a) pNext->next = a;
else pNext->next = b;
return head;
}

void MergeSort(Node **head)
{
if (!(*head) || !(*head)->next) return;
Node *a = NULL, *b = NULL;
FrontBackSplit(*head, &a, &b);
MergeSort(&a);
MergeSort(&b);
*head = SortedMerge(a, b);
}

//main函数进行测试
int main()
{
Node *n1 = new Node(2);
Node *n2 = new Node(1);
Node *n3 = new Node(4);
Node *n4 = new Node(3);
Node *n5 = new Node(5);
n1->next = n2;
n2->next = n3;
n3->next = n4;
n4->next = n5;
MergeSort(&n1);
while (n1)
{
cout << n1->val << " ";
n1 = n1->next;
}
cout << endl;
return 0;
}