链表指定值清除、链表的回文结构

你的名字 2022-06-09 03:19 144阅读 0赞

### 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 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;
        }
    }