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;
    }
};