力扣 Hot 100 链表算法题 Python 实现

一、反转链表相关

1.1 206. 反转链表

• 代码：

class ListNode:
    def __init__(self, val=0, next=None):
        self.val = val
        self.next = next

class Solution:
    def reverseList(self, head):
        pre = None
        cur = head
        while cur:
            nxt = cur.next
            cur.next = pre
            pre = cur
            cur = nxt
        return pre

1.2 92. 反转链表 II

• 思路：整体来看，left 是反转前的第 left 个节点（p0.next），pre 是反转后的头节点，cur 是当前遍历到的节点下一个。

class Solution:
    def reverseBetween(self, head, left, right):
        p0 = dummy = ListNode(next=head)
        # 这样一起实例化，p0 和 dummy 永远都是指向同一位置
        for _ in range(left - 1):
            p0 = p0.next
        # 知道转换的前一个结点
        pre = None
        cur = p0.next
        # 当前正在处理的节点
        for _ in range(right - left + ):
            nxt = cur.
            cur. = pre
            pre = cur
            cur = nxt
        p0.. = cur
        p0. = pre
         dummy.