力扣算法题—148sort-list

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

Example 1:

Input: 4->2->1->3
Output: 1->2->3->4

Example 2:

Input: -1->5->3->4->0
Output: -1->0->3->4->5

Solution:
　　复杂度为O(nlogn)有快速排序，并归排序，堆排序，对于来列表而言，堆排序最适合了
　　使用快慢指针将链表分为两部分
　　merge01为简洁版

 class Solution {
 public:
     ListNode* sortList(ListNode* head) {
         if (head == nullptr || head->next == nullptr)return head;
         ListNode *slow = head, *fast = head, *pre = head;
         while (fast != nullptr && fast->next != nullptr)
         {
             pre = slow;
             slow = slow->next;
             fast = fast->next->next;
         }
         pre->next = nullptr;//将链表分为两段
         return merge(sortList(head), sortList(slow));
     }
     ListNode* merge01(ListNode* l1, ListNode* l2)
     {
         if (l1 == nullptr || l2 == nullptr)return l1 == nullptr ? l2 : l1;
         if (l1->val < l2->val)
         {
             l1->next = merge(l1->next, l2);
             return l1;
         }
         else
         {
             l2->next = merge(l1, l2->next);
             return l2;
         }
     }
     ListNode* merge02(ListNode* l1, ListNode* l2)
     {
         ListNode* ptr = new ListNode(-);
         ListNode* p = ptr;
         while (l1 != nullptr && l2 != nullptr)
         {
             if (l1->val < l2->val)
             {
                 p->next = l1;
                 l1 = l1->next;
             }
             else
             {
                 p->next = l2;
                 l2 = l2->next;
             }
             p = p->next;
         }
         if (l1 == nullptr)p->next = l2;
         if (l2 == nullptr)p->next = l1;
         return ptr->next;
     }
 };