【LeetCode】每日一题：K个一组反转链表

孑渡2024-06-29 5:04

解题思路

其实更像一个模拟题，但是有两个地方的边界一直没有处理好导致卡了很久。

AC代码

python 复制代码

# 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]:
        def rotate(prev, tail):
        	# 少了一个end最开始，tail这个节点在循环修改next时候会变，从而导致循环终止条件错误，应该用一个定量记下来
            end = tail.next
            last = tail.next
            curr = prev.next
            res = prev.next
            while curr != end:
                temp = curr.next
                curr.next = last
                last = curr
                curr = temp
            prev.next = tail
            return res

        dummyhead = ListNode(0)
        dummyhead.next = head

        tail = dummyhead
        indexhead = dummyhead
        count = 0
    
        while indexhead:
            for _ in range(k):
            # 这个地方最开始写的是tail，其实应该是判断后面有没有了，后面没有就不能继续计数了
                if tail.next:
                    tail = tail.next
                    count += 1
                else:
                    break
            if count == k:
                tail = indexhead = rotate(indexhead, tail)
                count = 0
            else:
                break


        

        return dummyhead.next

官方题解

官方题解单独考虑了一个反转链表，然后把头尾返回，相当于把这个链表插回去

python 复制代码

class Solution:
    # 翻转一个子链表，并且返回新的头与尾
    def reverse(self, head: ListNode, tail: ListNode):
        prev = tail.next
        p = head
        while prev != tail:
            nex = p.next
            p.next = prev
            prev = p
            p = nex
        return tail, head

    def reverseKGroup(self, head: ListNode, k: int) -> ListNode:
        hair = ListNode(0)
        hair.next = head
        pre = hair

        while head:
            tail = pre
            # 查看剩余部分长度是否大于等于 k
            for i in range(k):
                tail = tail.next
                if not tail:
                    return hair.next
            nex = tail.next
            head, tail = self.reverse(head, tail)
            # 把子链表重新接回原链表
            pre.next = head
            tail.next = nex
            pre = tail
            head = tail.next
        
        return hair.next

作者：力扣官方题解
链接：https://leetcode.cn/problems/reverse-nodes-in-k-group/solutions/248591/k-ge-yi-zu-fan-zhuan-lian-biao-by-leetcode-solutio/
来源：力扣（LeetCode）
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