链表指定值清除

现在有一个单链表。链表中每个节点保存一个整数，再给定一个值val，把所有等于val的节点删掉。给定一个单链表的头结点head，同时给定一个值val，请返回清除后的链表的头结点，保证链表中有不等于该值的其它值。请保证其他元素的相对顺序。

测试样例：

{1,2,3,4,3,2,1},2

{1,3,4,3,1}我的提交

-- coding:utf-8 --

class ListNode:

def init(self, x):

self.val = x

self.next = None

class ClearValue:

def clear(self, head, val):

write code here

if not head:        return None    while head and head.val == val:        head = head.next    p = head    h = None    while p:        if p.val == val:            h.next = p.next        else:            h = p        p = p.next    #h.next = None    return head

链表的回文结构

请编写一个函数，检查链表是否为回文。给定一个链表ListNode* pHead，请返回一个bool，代表链表是否为回文。#测试样例：{1,2,3,2,1}#返回：true{1,2,3,2,3}#返回：false我的提交

-- coding:utf-8 --

class ListNode:

def init(self, x):

self.val = x

self.next = None

class Palindrome:

def isPalindrome(self, pHead):

write code here

stack = []    p = pHead    while p:        temp = p        stack.append(temp)        p = p.next    p = pHead    for _ in range((len(stack) // 2) + 1):        temp = stack.pop()        if p.val != temp.val:            return False        p = p.next    return True

参考答案

import java.util.*;

/*

public class ListNode { int val;ListNode next = null;

ListNode(int val) {    this.val = val;}

}*/

public class Palindrome { public boolean isPalindrome(ListNode head) { if (head == null || head.next == null) { return true;}ListNode n1 = head;ListNode n2 = head;while (n2.next != null && n2.next.next != null) { // find mid noden1 = n1.next; // n1 -> midn2 = n2.next.next; // n2 -> end}n2 = n1.next; // n2 -> right part first noden1.next = null; // mid.next -> nullListNode n3 = null;while (n2 != null) { // right part convertn3 = n2.next; // n3 -> save next noden2.next = n1; // next of right node convertn1 = n2; // n1 moven2 = n3; // n2 move}n3 = n1; // n3 -> save last noden2 = head;// n2 -> left first nodeboolean res = true;while (n1 != null && n2 != null) { // check palindromeif (n1.val != n2.val) { res = false;break;}n1 = n1.next; // left to midn2 = n2.next; // right to mid}n1 = n3.next;n3.next = null;while (n1 != null) { // recover listn2 = n1.next;n1.next = n3;n3 = n1;n1 = n2;}return res;}}