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

因为题目要求复杂度为O(nlogn),故可以考虑归并排序的思想。

归并排序的一般步骤为：

1）将待排序数组（链表）取中点并一分为二；

2）递归地对左半部分进行归并排序；

3）递归地对右半部分进行归并排序；

4）将两个半部分进行合并（merge）,得到结果。

 

所以对应此题目，可以划分为三个小问题：

1）找到链表中点 （快慢指针思路，快指针一次走两步，慢指针一次走一步，快指针在链表末尾时，慢指针恰好在链表中点）；

2）写出merge函数，即如何合并链表。 （见merge-two-sorted-lists 一题解析）

3）写出mergesort函数，实现上述步骤

class Solution {
public:
    ListNode* findMiddle(ListNode* head){
        ListNode* chaser = head;
        ListNode* runner = head->next;
        while(runner != NULL && runner->next != NULL){
            chaser = chaser->next;
            runner = runner->next->next;
        }
        return chaser;
    }
     
 ListNode* mergeTwoLists(ListNode* l1, ListNode* l2) {
        if(l1 == NULL){
            return l2;
        }
        if(l2 == NULL){
            return l1;
        }
        ListNode* dummy = new ListNode(0);
        ListNode* head = dummy;
        while(l1 != NULL && l2 != NULL){
            if(l1->val > l2->val){
                head->next = l2;
                l2 = l2->next;
            }
            else{
                head->next = l1;
                l1 = l1->next;
            }
            head = head->next;
        }
        if(l1 == NULL){
            head ->next = l2;
        }
        if(l2 == NULL){
            head->next = l1;
        }
        return dummy->next;
    }
     
    ListNode* sortList(ListNode* head) {
        if(head == NULL || head ->next == NULL){
            return head;
        }
        ListNode* middle = findMiddle(head);
        ListNode* right = sortList(middle->next);
        middle -> next = NULL;
        ListNode* left = sortList(head);
        return mergeTwoLists(left, right);
    }
};