09-07 leetcode-0092

题目

Given the head of a singly linked list and two integers left and right where left <= right, reverse the nodes of the list from position left to position right, and return the reversed list.

题解

from typing import Optional


class ListNode:
    def __init__(self, val: int = 0, next: Optional["ListNode"] = None):
        self.val = val
        self.next = next


class Solution:
    def reverseBetween(
        self, head: Optional["ListNode"], left: int, right: int
    ) -> Optional["ListNode"]:
        if not head or left >= right:
            return head
        dummy = ListNode(-1, head)
        start, end = dummy, dummy
        left_node = None
        count = 1
        convert_flag = False
        while head:
            if not convert_flag:
                if count == left:
                    start = end
                    end, left_node = head, head
                    convert_flag = True
                else:
                    end = head
                head = head.next
            else:
                temp = head.next
                if count == right:
                    head.next = end
                    left_node.next = temp
                    if start:
                        start.next = head
                    break
                else:
                    head.next = end
                    end = head
                    head = temp
            count += 1
        return dummy.next

我写会这道题，是不是可以去 Google 了（

Comments