leetcode_234题——Palindrome Linked List（链表）

Palindrome Linked List

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Question  Solution 

 

Given a singly linked list, determine if it is a palindrome.

Follow up:
Could you do it in O(n) time and O(1) space?

 

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  (E) Palindrome Number (E) Valid Palindrome (E) Reverse Linked List

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这题需要O（n）的计算时间复杂度，需要O（1）的存储空间，所以可以将后半部分链表逆序，再一个个比较

#include<iostream>

#include<vector>

using namespace std;



struct ListNode {

      int val;

      ListNode *next;

      ListNode(int x) : val(x), next(NULL) {}

 };



ListNode* reversed(ListNode* L1,ListNode* L2)

{

	if(L1==L2)

	{

		L1->next==NULL;

		return L1;

	}

	if(L1->next==L2)

	{

		L2->next=L1;

		L1->next=NULL;

		return L2;

	}

	ListNode* ptr1=L1;

	ListNode* ptr2=ptr1->next;

	ListNode* ptr3=ptr2->next;

	while(ptr3!=L2)

	{

		ptr2->next=ptr1;

		ptr1=ptr2;

		ptr2=ptr3;

		ptr3=ptr3->next;

	}

	ptr2->next=ptr1;

	ptr1->next=NULL;

	return ptr2;

}



bool isPalindrome(ListNode* head) {

	if(head==NULL)

		return 0;

	if(head->next==NULL)

		return 1;

	ListNode* ptr1=head;

	int n=1;

	while(1)

	{

		if(ptr1->next==NULL)

			break;

		n++;

		ptr1=ptr1->next;

	}



	int mid=n/2;

	ListNode* ptr2=head;

	int jishu=1;

	while(1)

	{

		if(jishu==mid)

			break;

		jishu++;

		ptr2=ptr2->next;

	}



	ListNode* ptr3=ptr2->next;

	ptr1=reversed(ptr3,ptr1);



	ListNode* ptr0=head;

	if(n/2==0)

	{

		while(ptr1!=NULL)

		{

			if(ptr1->next!=ptr0->next)

				return 0;

			ptr1=ptr1->next;

			ptr0=ptr0->next;

		}

		return 1;

	}

	else

	{

		while(ptr1!=ptr3)

		{

			if(ptr1->next!=ptr0->next)

				return 0;

			ptr1=ptr1->next;

			ptr0=ptr0->next;

		}

		return 1;

	}

}





int main()

{

	int vec[20]={-31900,22571,-31634,19735,13748,16612,-28299,-16628,9614,-14444,

		      -14444,9614,-16628,-31900,16612,13748,19735,-31634,22571,-28299};

	ListNode* root=new ListNode(-31900);

	ListNode* ptr1=root;

	int i=1;

	while(i<20)

	{

		ptr1->next=new ListNode(vec[i]);

		ptr1=ptr1->next;		

		i++;

	}

	cout<<isPalindrome(root)<<endl;

}