Leetcode刷题Python92.反转链表II

Posted 2022-05-16 Better Bench

1 题目

给你单链表的头指针 head 和两个整数 left 和 right ，其中 left <= right 。请你反转从位置 left 到位置 right 的链表节点，返回 反转后的链表 。

2 解析

（1）第一种：头插法

（2）第二种：迭代法，先截取中间部分，然后反转，再连接两头

（3）第三种：递归法
待更新

3 Python实现

class Solution:
    # 方法一：头插法
    '''
    def reverseBetween(self, head: ListNode, left: int, right: int) -> ListNode:
        dummy = ListNode(-1)
        dummy.next = head
        prev = dummy
        for _ in range(left-1):
            prev = prev.next
        curr = prev.next
        for _ in range(right-left):
            next_node = curr.next
            curr = next_node.next
            next_node.next = prev.next
            prev.next = next_node
        return dummy
    '''
    # 方法二：迭代法
    def reverseBetween(self, head: ListNode, left: int, right: int) -> ListNode:
        
        def reverseList(head: ListNode):
            pre = None
            cur = head
            while cur:
                next = cur.next
                cur.next = pre
                pre = cur
                cur = next
        dummy_node = ListNode(-1)
        dummy_node.next = head
        prev  = dummy_node

        for _ in range(left-1):
            prev = prev.next
        right_node = prev
        for _ in range(right-left+1):
            right_node = right_node.next
        # 保存节点
        left_node = prev.next
        curr = right_node.next
        # 切断链表
        prev.next = None
        right_node.next = None

        reverseList(left_node)
        # 恢复连接
        # 注意，此处右node变成了左node，左node变成了右node。
        prev.next = right_node
        left_node.next = curr
        return  dummy_node.next