链表

链表中每一个节点为一个对象，对象中包含两个成员变量，第一个是val，代表链表的值，第二个是next，它指向下一个节点，是下一个节点对象的引用。

定义链表

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

删除节点

给定链表中的某个节点，删除该节点

class Solution {public void deleteNode(ListNode node) {node.val = node.next.val;node.next = node.next.next;}
}

反转链表

普通方法：

def reverseList(head: Optional[ListNode]) -> Optional[ListNode]:newHead = Nonewhile head != None:temp = head.nexthead.next = newHeadnewHead = headhead = tempreturn newHead

使用栈解决:
一定要注意让尾巴节点next指针为空，否则将形成环：

def reverseList(head: Optional[ListNode]) -> Optional[ListNode]:if head == None:return Nonea = []while head != None:a.append(head)head = head.nexthead = a.pop()b = headwhile len(a) > 0:n = a.pop()b.next = nb = nb.next = Nonereturn head

合并两个有序链表

合并两个升序链表，变成一个新的升序链表

递归：

def mergeTwoLists(self, list1: Optional[ListNode], list2: Optional[ListNode]) -> Optional[ListNode]:if list1 is None: return list2if list2 is None: return list1if list1.val <= list2.val:list1.next = self.mergeTwoLists(list1.next, list2)return list1else:list2.next = self.mergeTwoLists(list1,list2.next)return list2

判断链表是否有环

直观解法：

class Solution:def hasCycle(self, head: Optional[ListNode]) -> bool:a = []while head != None:if head not in a:a.append(head)head = head.nextelse:return Truereturn False

快慢指针：

class Solution:def hasCycle(self, head: Optional[ListNode]) -> bool:head1 = headhead2 = headwhile head1 != None and head2 != None and head2.next != None:head1 = head1.nexthead2 = head2.next.nextif head1 == head2:return Truereturn False