链表算法实战：反转、合并与排序（Python）

1.3 K 个一组翻转链表

Python 代码

class Solution:
    def reverseKGroup(self, head: Optional[ListNode], k: int) -> Optional[ListNode]:
        n = 0
        cur = head
        while cur:
            n += 1
            cur = cur.next
        
        p0 = dummy = ListNode(next=head)
        pre = None
        cur = head
        
        while n >= k:
            n -= k
            for _ in range(k):
                nxt = cur.next
                cur.next = pre
                pre = cur
                cur = nxt
            nxt = p0.next
            nxt.next = cur
            p0.next = pre
            p0 = nxt
        return dummy.next

二、链表排序

2.1 链表的中间结点

思路

典型的快慢指针法。t1 每次走一步，t2 每次走两步，当 t2 到达末尾时，t1 即为中间节点。

Python 代码

class Solution:
    def middleNode(self, head: Optional[ListNode]) -> Optional[ListNode]:
        t1 = t2 = head
        while t2 and t2.next:
            t1 = t1.next
            t2 = t2.next.next
        return t1

2.2 合并两个有序链表

Python 代码

class Solution:
    def mergeTwoLists(self, list1: Optional[ListNode], list2: Optional[ListNode]) -> Optional[ListNode]:
        cur = dummy = ListNode()
        while list1 and list2:
            if list1.val <= list2.val:
                cur.next = list1
                cur = cur.next
                list1 = list1.next
            else:
                cur.next = list2
                cur = cur.next
                list2 = list2.next
        cur.next = list1 if list1 else list2
        return dummy.next

2.3 排序链表

思路

归并排序。找到链表的中间节点，断开为前后两端，分别排序前后两端，排序后再合并两个有序链表。

Python 代码

class Solution:
    # 876. 链表的中间结点（快慢指针）
    def middleNode(self, head: Optional[ListNode]) -> Optional[ListNode]:
        slow = fast = head
        while fast and fast.next:
            pre = slow
            slow = slow.next
            fast = fast.next.next
        pre.next = None
        return slow

    # 21. 合并两个有序链表（双指针）
    def mergeTwoLists(self, list1: Optional[ListNode], list2: Optional[ListNode]) -> Optional[ListNode]:
        cur = dummy = ListNode()
        while list1 and list2:
            if list1.val < list2.val:
                cur.next = list1
                list1 = list1.next
            else:
                cur.next = list2
                list2 = list2.next
            cur = cur.next
        cur.next = list1 if list1 else list2
        return dummy.next

    def sortList(self, head: Optional[ListNode]) -> Optional[ListNode]:
        if head is None or head.next is None:
            return head
        head2 = self.middleNode(head)
        head = self.sortList(head)
        head2 = self.sortList(head2)
        return self.mergeTwoLists(head, head2)