pengweixuan008的博客

题目：Remove all elements from a linked list of integers that have value val.ExampleGiven: 1 --> 2 --> 6 --> 3 --> 4 --> 5 --> 6, val = 6Return: 1 --> 2 --> 3 --> 4 --> 5思路：新建一个val=-1的头节点，从he

Reverse a singly linked list./** * Definition for singly-linked list. * public class ListNode { *     int val; *     ListNode next; *     ListNode(int x) { val = x; } * }   **/  方法

1：由于要求链表是有序的，可以借助二叉树中序遍历，因为中序遍历算法的特点就是从小到大访问结点。当遍历访问到根结点时，假设根结点的左侧已经处理好，只需将根结点与上次访问的最近结点（左子树中最大值结点）的指针连接好即可。进而更新当前链表的最后一个结点指针。2：由于中序遍历过程正好是转换成链表的过程，即可采用递归处理struct BinaryTreeNode  {

public class Solution {    public boolean hasCycle(ListNode head) {     ListNode slow = head, fast = head;          while (fast != null && fast.next != null) {        slow = slow.next;   

Write a program to find the node at which the intersection of two singly linked lists begins.For example, the following two linked lists:A: a1 → a2 ↘

For example,Given 1->1->2, return 1->2.Given 1->1->2->3->3, return 1->2->3.public class Solution {    public ListNode deleteDuplicates(ListNode head) {        if(head==null) return n

Given a linked list, remove the nth node from the end of list and return its head.For example, Given linked list: 1->2->3->4->5, and n = 2. After removing the second node from the end,