Sort list by merge sort

使用归并排序对链表进行排序

O(nlgn) 的时间效率

/**

 * Definition for singly-linked list.

 * struct ListNode {

 *     int val;

 *     ListNode *next;

 *     ListNode(int x) : val(x), next(NULL) {}

 * };

 */

class Solution {

public:

    ListNode* sortList(ListNode* head) {

        if (head == NULL || head->next == NULL) return head;    //one node or none return head

        

        ListNode * fast = head;                                 //to find the mid by using fast and slow pointer

        ListNode * slow = head;

        

        while (fast->next != NULL && fast->next->next != NULL) {

            fast = fast->next->next;

            slow = slow->next;

        }

        

        fast = slow->next;

        slow->next = NULL;

        

        ListNode * l1 = sortList (head);                        //merge sort in O(nlgn)

        ListNode * l2 = sortList (fast);

        

        return merge (l1, l2);

    }

    

    ListNode * merge (ListNode * l1, ListNode * l2) {

        ListNode * head = new ListNode (-1);

        ListNode * p = head;

        

        while (l1 != NULL && l2 != NULL) {

            if (l1->val < l2->val) {

                p->next = l1;

                l1 = l1->next;

            } else {

                p->next = l2;

                l2 = l2->next;

            }

            p = p->next;

        }

        

        p->next = (l1 == NULL) ? l2 : l1;

        

        return head->next;

    }

};