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