题目浅析

  • 想查看原题可以点击题目链接。

  • 简单地说，就是给一个链表和一个数字 k，现在要从左到右，每 k 个就翻转链表。

思路分享

代码解答（强烈建议自行解答后再看）

  • 参考题解
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# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution:
    def reverseKGroup(self, head: Optional[ListNode], k: int) -> Optional[ListNode]:
        n = 0
        cur = head
        while cur:
            n += 1
            cur = cur.next
        
        m = n // k
        dummy = ListNode(0, head)
        p0 = dummy
        for _ in range(m):
            pre = p0
            cur = pre.next
            for i in range(k):
                nxt = cur.next
                cur.next = pre
                pre = cur
                cur = nxt
            nxt = p0.next
            p0.next.next = cur
            p0.next = pre
            p0 = nxt

        return dummy.next