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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Linked List Two Pointers

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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;
}