Leetcode 234. Palindrome Linked List

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234. Palindrome Linked List

Total Accepted: 69120 Total Submissions: 225579 Difficulty: Easy

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

思路：

解法一：

n空间的做法是遍历到中间之后把slow.next开始的所有元素入栈。无论奇偶slow.next永远是要做对比的元素。然后做对比。

/** * Definition for singly-linked list. * public class ListNode { *     int val; *     ListNode next; *     ListNode(int x) { val = x; } * } */public class Solution { // 6ms    public boolean isPalindrome(ListNode head) {        if(head == null) return true;        ListNode slow = head;        ListNode fast = head;        while(fast.next != null && fast.next.next!=null){            slow = slow.next;            fast = fast.next.next;        }                Stack<Integer> st = new Stack<>();                fast = slow.next;        while(fast != null){            st.push(fast.val);            fast = fast.next;        }        while(!st.isEmpty()){            if(st.pop() != head.val) return false;            head = head.next;        }        return true;    }}

解法二：

常数空间。从以slow作为dummy head，把之后的元素翻转。然后slow = slow.next 在slow != null的时候和head元素对比。

/** * Definition for singly-linked list. * public class ListNode { *     int val; *     ListNode next; *     ListNode(int x) { val = x; } * } */public class Solution { // 2ms    public boolean isPalindrome(ListNode head) {        if(head == null) return true;        ListNode slow = head;        ListNode fast = head;        while(fast.next != null && fast.next.next!=null){            slow = slow.next;            fast = fast.next.next;        }        fast = slow.next; slow.next = null;        while(fast != null){            ListNode next = fast.next;            fast.next = slow.next;            slow.next = fast;            fast = next;        }                slow = slow.next;        while(slow !=null){            if(slow.val != head.val) return false;            slow = slow.next;            head = head.next;        }        return true;    }}

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