LeetCode148：Sort List

题目：

Sort a linked list in O(n log n) time using constant space complexity.

解题思路：

根据题目要求，可知只能用归并排序，其他排序算法要么时间复杂度不满足，要么空间复杂度不满足

实现代码：

#include <iostream>

using namespace std;
/*
Sort a linked list in O(n log n) time using constant space complexity.
*/
/**
* Definition for singly-linked list.
* struct ListNode {
*     int val;
*     ListNode *next;
*     ListNode(int x) : val(x), next(NULL) {}
* };
*/
struct ListNode
{
int val;
ListNode *next;
ListNode(int x):val(x), next(NULL)
{

}

};

void addNode(ListNode* &head, int val)
{
ListNode *newNode = new ListNode(val);
if(head == NULL)
{
head = newNode;
}
else
{
newNode->next = head;
head = newNode;
}
}

void PrintList(ListNode *root)
{
ListNode *head = root;
while(head != NULL)
{
cout<<head->val<<"\t";
head = head->next;
}
cout<<endl;
}

class Solution {
public:
ListNode *sortList(ListNode *head) {
if(head == NULL || head->next == NULL)
return head;
ListNode *quick = head;
ListNode *slow = head;
while(quick->next && quick->next->next)//通过两个指针，一个走两步、一个走一步，获得链表的中间节点
{
slow = slow->next;
quick = quick->next->next;
}
quick = slow;
slow = slow->next;
quick->next = NULL;//将链表的前半段进行截断
ListNode *head1 = sortList(head);
ListNode *head2 = sortList(slow);
return merge(head1, head2);
}

//归并两个有序链表
ListNode *merge(ListNode *head1, ListNode *head2)
{
if(head1 == NULL)
return head2;
if(head2 == NULL)
return head1;
ListNode *newHead = NULL;
if(head1->val < head2->val)
{
newHead = head1;
head1 = head1->next;

}
else
{
newHead = head2;
head2 = head2->next;
}
ListNode *p = newHead;
while(head1 && head2)
{
if(head1->val < head2->val)
{
p->next = head1;
head1 = head1->next;
}
else
{
p->next = head2;
head2 = head2->next;
}
p = p->next;
}
if(head1)
p->next = head1;
if(head2)
p->next = head2;
return newHead;
}
};

int main(void)
{
ListNode *head = new ListNode(5);
addNode(head, 3);
addNode(head, 10);
addNode(head, 15);
PrintList(head);

Solution solution;
head = solution.sortList(head);
PrintList(head);

return 0;
}