链表指定值清除、链表的回文结构

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 = Noneclass 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

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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 = Noneclass 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 node            n1 = n1.next; // n1 -> mid            n2 = n2.next.next; // n2 -> end        }        n2 = n1.next; // n2 -> right part first node        n1.next = null; // mid.next -> null        ListNode n3 = null;        while (n2 != null) { // right part convert            n3 = n2.next; // n3 -> save next node            n2.next = n1; // next of right node convert            n1 = n2; // n1 move            n2 = n3; // n2 move        }        n3 = n1; // n3 -> save last node        n2 = head;// n2 -> left first node        boolean res = true;        while (n1 != null && n2 != null) { // check palindrome            if (n1.val != n2.val) {                res = false;                break;            }            n1 = n1.next; // left to mid            n2 = n2.next; // right to mid        }        n1 = n3.next;        n3.next = null;        while (n1 != null) { // recover list            n2 = n1.next;            n1.next = n3;            n3 = n1;            n1 = n2;        }        return res;    }}