[leetcode]Sort List

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

数据结构定义：

1 /**
2  * Definition for singly-linked list.
3  * struct ListNode {
4  *     int val;
5  *     ListNode *next;
6  *     ListNode(int x) : val(x), next(NULL) {}
7  * };
8  */

这里用的数据结构是单链表。时间复杂度是O(n log n)的排序算法有：堆排序，快速排序，归并排序。题目要求使用常量的空间复杂度，这三种排序方法的空间复杂度分别是O(n),O(1),O(n)[对于数组]。归并排序的空间复杂度是O(n)，因为对于数组，归并排序需要额外的O(n)的空间来处理数组，而对于链表而言，操作都是对指针的，所以可以将空间复杂度降到O(1)。

这里使用归并排序来实现代码。

 1 /**
 2  * Definition for singly-linked list.
 3  * struct ListNode {
 4  *     int val;
 5  *     ListNode *next;
 6  *     ListNode(int x) : val(x), next(NULL) {}
 7  * };
 8  */
 9 class Solution {
10 public:
11     ListNode *sortList(ListNode *head){
12         if (head == NULL)
13             return head;
14         else if (head->next == NULL) 
15             return head;
16         
17         ListNode *slow = head, *fast = head;
18         while (fast->next != NULL && fast->next->next != NULL)
19         {
20             slow = slow->next;
21             fast = fast->next->next;
22         }
23         
24         fast = slow;
25         slow = slow->next;
26         fast->next = NULL;
27         
28         fast = sortList(head);
29         slow = sortList(slow);
30         
31         return mergeList(fast,slow);
32         
33     }
34     ListNode* mergeList(ListNode* list1Head, ListNode* list2Head){
35         if (list1Head == NULL) return list2Head;
36         if (list2Head == NULL) return list1Head;
37         
38         ListNode* result, *temp;
39         // initialize the first node of the merged result list
40         if (list1Head->val < list2Head->val){
41             result = list1Head;
42             list1Head = list1Head->next;
43         }
44         else
45         {
46             result = list2Head;
47             list2Head = list2Head->next;
48         }
49         // operate the result list with temp
50         temp = result;
51         // end merge if at least one of the two lists reaches the tail node
52         while (list1Head != NULL && list2Head != NULL){
53             if (list1Head->val < list2Head->val){
54                 temp->next = list1Head;
55                 list1Head = list1Head->next;
56             }
57             else{
58                 temp->next = list2Head;
59                 list2Head = list2Head->next;
60             }
61             temp = temp->next;
62         }
63         
64         if (list1Head != NULL)
65             temp->next = list1Head;
66         else
67             temp->next = list2Head;
68             
69         return result;
70     }
71 };