234. Palindrome Linked List

题目：

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?

链接： http://leetcode.com/problems/palindrome-linked-list/

题解：

判断链表是否是Palindrome。 我们分三步解，先用快慢指针找中点，接下来reverse中点及中点后部，最后逐节点对比值。

Time Complexity - O(n)， Space Complexity - O(1)

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
public class Solution {
    public boolean isPalindrome(ListNode head) {
        if(head == null || head.next == null)
            return true;
        ListNode mid = findMid(head);
        ListNode tail = reverse(mid);
        mid.next = null;

        while(head != null && tail != null) {
            if(head.val != tail.val)
                return false;
            else {
                head = head.next;
                tail = tail.next;
            }
        }

        return true;
    }

    private ListNode findMid(ListNode head) {               // find mid node of list
        if(head == null || head.next == null)
            return head;
        ListNode slow = head, fast = head;

        while(fast != null && fast.next != null) {
            fast = fast.next.next;
            slow = slow.next;
        }

        return slow;
    }

    private ListNode reverse(ListNode head) {           // reverse listnode
        if(head == null || head.next == null)
            return head;
        ListNode dummy = new ListNode(-1);
        while(head != null) {
            ListNode tmp = head.next;
            head.next = dummy.next;
            dummy.next = head;
            head = tmp;
        }
        return dummy.next;
    }
}

二刷：

和一刷一样，先快慢指针找重点，然后reverse后半部分，接下来遍历两个head逐个对比节点的值。最后返回true.

Java

Time Complexity - O(n)， Space Complexity - O(1)

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
public class Solution {
    public boolean isPalindrome(ListNode head) {
        if (head == null || head.next == null) {
            return true;
        }
        ListNode mid = findMid(head);
        ListNode reversedMid = reverse(mid);
        ListNode node = head;
        while (node != null && reversedMid != null) {
            if (node.val != reversedMid.val) {
                return false;
            }
            node = node.next;
            reversedMid = reversedMid.next;
        }
        return true;
    }

    private ListNode findMid(ListNode head) {
        ListNode fast = head, slow = head;
        while (fast != null && fast.next != null) {
            fast = fast.next.next;
            slow = slow.next;
        }
        return slow;
    }

    private ListNode reverse(ListNode head) {
        ListNode dummy = new ListNode(-1);
        ListNode next = null;
        while (head != null) {
            next = head.next;
            head.next = dummy.next;
            dummy.next = head;
            head = next;
        }
        return dummy.next;
    }

}

三刷:

跟二刷一样

Java:

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
public class Solution {
    public boolean isPalindrome(ListNode head) {
        if (head == null || head.next == null) return true;
        ListNode mid = findMid(head);
        ListNode reversedMid = reverse(mid);
        ListNode node = head;
        while (node != null && reversedMid != null) {
            if (node.val != reversedMid.val) return false;
            node = node.next;
            reversedMid = reversedMid.next;
        }
        return true;
    }

    private ListNode findMid(ListNode head) {
        ListNode fast = head, slow = head;
        while (fast != null && fast.next != null) {
            fast = fast.next.next;
            slow = slow.next;
        }
        return slow;
    }

    private ListNode reverse(ListNode head) {
        ListNode dummy = new ListNode(-1);
        ListNode next = null;
        while (head != null) {
            next = head.next;
            head.next = dummy.next;
            dummy.next = head;
            head = next;
        }
        return dummy.next;
    }

}

Update:

这样写确实会破坏原来链表的结构。而且反转后半部分的时候是否可以可以算作O(1) space也值得商榷

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
public class Solution {
    public boolean isPalindrome(ListNode head) {
        if (head == null || head.next == null) return true;
        ListNode mid = findMid(head);
        ListNode tailReversed = reverse(mid);
        while (tailReversed != null && head != null) {
            if (tailReversed.val != head.val) return false;
            tailReversed = tailReversed.next;
            head = head.next;
        }
        return true;
    }

    private ListNode findMid(ListNode head) {
        ListNode fast = head, slow = head;
        while (fast != null && fast.next != null) {
            fast = fast.next.next;
            slow = slow.next;
        }
        return slow;
    }

    private ListNode reverse(ListNode head) {
        ListNode dummy = new ListNode(-1);
        ListNode tmp = null;
        while (head != null) {
            tmp = head.next;
            head.next = dummy.next;
            dummy.next = head;
            head = tmp;
        }
        return dummy.next;
    }
}

Reference：

https://leetcode.com/discuss/44751/11-lines-12-with-restore-o-n-time-o-1-space