#98 Sort List

题目描述：

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

Have you met this question in a real interview? 

Yes

Example

Given 1->3->2->null, sort it to 1->2->3->null.

Challenge 

Solve it by merge sort & quick sort separately.

题目思路：

我用了merge sort的算法，这里因为这个题是linked list，和array sort有所不同：需要找到list的中间点，把list均匀分成两部分（注意要把中间点断掉）。找到以后就各自mergesort一番，再把两个sorted lists merge一下就好了（这里merge也用了好用的dummy head，就不用纠结merge好的list用list1做head还是list2做head了）。

Mycode（AC = 60ms）：

/** * Definition of ListNode * class ListNode { * public: *     int val; *     ListNode *next; *     ListNode(int val) { *         this->val = val; *         this->next = NULL; *     } * } */class Solution {public:    /**     * @param head: The first node of linked list.     * @return: You should return the head of the sorted linked list,                    using constant space complexity.     */    ListNode *sortList(ListNode *head) {        // write your code here        return mergeSortList(head);    }        ListNode *mergeSortList(ListNode *head) {        if (head == NULL || head->next == NULL) {            return head;        }                // find the middle of list        ListNode *slow = head, *fast = head;        while (fast && fast->next && fast->next->next) {            slow = slow->next;            fast = fast->next->next;        }                // evenly break the list into two lists        ListNode *head2 = slow->next;        slow->next = NULL;                ListNode *list1 = mergeSortList(head);        ListNode *list2 = mergeSortList(head2);        return merge(list1, list2); // merge two sorted lists    }        // merge two sorted lists    ListNode *merge(ListNode *head1, ListNode *head2) {        ListNode *dummy = new ListNode(0);        ListNode *tmp = dummy, *h1 = head1, *h2 = head2;                while (h1 && h2) {            if (h1->val <= h2->val) {                tmp->next = h1;                h1 = h1->next;                tmp = tmp->next;            }            else {                tmp->next = h2;                h2 = h2->next;                tmp = tmp->next;            }        }                if (h1) tmp->next = h1;        if (h2) tmp->next = h2;                return dummy->next;    }};