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; } }
