腾讯精选50题（6）

148.排序链表
https://leetcode-cn.com/problems/sort-list/
在 O(n log n) 时间复杂度和常数级空间复杂度下，对链表进行排序。

示例 1:

输入: 4->2->1->3
输出: 1->2->3->4

示例 2:

输入: -1->5->3->4->0
输出: -1->0->3->4->5

分析：
先拆分后合并，我觉得这题有点难，直接看题解更容易懂

C++

class Solution {
public:
   
    ListNode* sortList(ListNode* head) {
         ListNode *fast,*slow,*mid,*left,*right;
        if(head==NULL||head->next==NULL) return head;
        slow = head;
        fast = head->next;
        while(fast&&fast->next){
            fast = fast->next->next;
            slow = slow->next;
        }
        mid = slow->next;
        slow->next = NULL;
        left = sortList(head);
        right = sortList(mid);
        ListNode* res = new ListNode(0);
        ListNode* h = res;
        while(left&&right){
            if(left->val<right->val){
                h->next = left;
                left = left->next;
            }
            else{
                h->next=right;
                right = right->next;
            }
            h=h->next;
        }
        if(left) h->next=left;
        if(right) h->next=right;
        return res->next;
    }
};

Python3

class Solution:
    def sortList(self, head: ListNode) -> ListNode:
        if not head or not head.next:
            return head
        fast,slow = head.next,head
        while fast and fast.next:
            fast = fast.next.next
            slow = slow.next
        mid = slow.next
        slow.next = None
        left = self.sortList(head)
        right = self.sortList(mid)
        h = res = ListNode(0)
        while left and right:
            if left.val<right.val:
                h.next=left
                left = left.next
            else:
                h.next=right
                right = right.next
            h = h.next
        if left:
            h.next = left
        if right:
            h.next = right
        return res.next