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‎README.md

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‎README_EN.md

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‎index.html

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<meta name="viewport" content="width=device-width, user-scalable=no, initial-scale=1.0, maximum-scale=1.0, minimum-scale=1.0">
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<link rel="stylesheet" href="//unpkg.com/docsify/lib/themes/vue.css">
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<link rel="stylesheet" href="//unpkg.com/[email protected]/dist/style.css" />
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<link rel="icon" type="image/png" sizes="32x32" href="img/favicon-32x32.png">
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<link rel="icon" type="image/png" sizes="16x16" href="img/favicon-16x16.png">
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<link rel="icon" type="image/png" sizes="32x32" href="images/favicon-32x32.png">
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<link rel="icon" type="image/png" sizes="16x16" href="images/favicon-16x16.png">
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</head>
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<body>
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<div id="app">本系列知识由 Doocs 开源社区原创发布</div>
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<div id="app">本系列知识由 Doocs 技术社区原创发布</div>
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<script>
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window.$docsify = {
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name: 'leetcode',
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logo: '/img/doocs-leetcode.png',
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logo: '/images/doocs-leetcode.png',
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search: [
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'/','/solution/','/lcof/','lcci/'
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],

‎lcof/面试题18. 删除链表的节点/README.md

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‎lcof/面试题18. 删除链表的节点/Solution.java

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/**
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* Definition for singly-linked list. public class ListNode { int val; ListNode
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* next; ListNode(int x) { val = x; } }
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* Definition for singly-linked list.
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* public class ListNode {
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* int val;
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* ListNode next;
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* ListNode(int x) { val = x; }
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* }
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*/
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class Solution {
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public ListNode deleteNode(ListNode head, int val) {
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ListNode pre = new ListNode(0);
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pre.next = head;
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ListNode dummy = pre, p = head;
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ListNode dummy = new ListNode(0);
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dummy.next = head;
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ListNode pre = dummy, p = head;
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while (p != null) {
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if (p.val == val) {
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pre.next = p.next;
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break;
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} else {
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pre = p;
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p = p.next;
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}
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pre = p;
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p = p.next;
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}
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return dummy.next;
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}

‎lcof/面试题18. 删除链表的节点/Solution.py

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# def __init__(self, x):
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# self.val = x
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# self.next = None
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class Solution:
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def deleteNode(self, head: ListNode, val: int) -> ListNode:
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pre = ListNode(0)
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pre.next = head
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dummy = pre
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p = head
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dummy = ListNode(0)
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dummy.next = head
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pre, p = dummy, head
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while p:
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if p.val == val:
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pre.next = p.next
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break
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else:
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pre, p = p, p.next
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pre, p = p, p.next
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return dummy.next

‎lcof/面试题22. 链表中倒数第k个节点/README.md

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‎lcof/面试题24. 反转链表/README.md

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‎lcof/面试题24. 反转链表/Solution.java

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*/
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class Solution {
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public ListNode reverseList(ListNode head) {
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if (head == null) {
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return head;
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}
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ListNode dummy = new ListNode(0);
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ListNode pre = null;
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ListNode p = head;
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while (p != null) {
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ListNode q = p.next;
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p.next = dummy.next;
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dummy.next = p;
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p.next = pre;
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pre = p;
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p = q;
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}
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return dummy.next;
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return pre;
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}
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}

‎lcof/面试题24. 反转链表/Solution.py

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class Solution:
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def reverseList(self, head: ListNode) -> ListNode:
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if not head:
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return head
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dummy = ListNode(0)
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p = head
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pre, p = None, head
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while p:
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q = p.next
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p.next = dummy.next
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dummy.next = p
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p.next = pre
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pre = p
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p = q
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return dummy.next
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return pre

‎lcof/面试题25. 合并两个排序的链表/README.md

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‎lcof/面试题35. 复杂链表的复制/README.md

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‎lcof/面试题35. 复杂链表的复制/images/e1.png

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‎lcof/面试题35. 复杂链表的复制/images/e2.png

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‎lcof/面试题36. 二叉搜索树与双向链表/README.md

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‎lcof/面试题36. 二叉搜索树与双向链表/images/bstdlloriginalbst.png

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‎lcof/面试题36. 二叉搜索树与双向链表/images/bstdllreturndll.png

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‎lcof/面试题52. 两个链表的第一个公共节点/README.md

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‎lcof/面试题52. 两个链表的第一个公共节点/images/160_example_1.png

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‎lcof/面试题52. 两个链表的第一个公共节点/images/160_example_2.png

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‎lcof/面试题52. 两个链表的第一个公共节点/images/160_example_3.png

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‎lcof/面试题52. 两个链表的第一个公共节点/images/160_statement.png

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