leetcode234 - Palindrome Linked List - easy

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

Example 1:

Input: 1->2

Output: false

Example 2:

Input: 1->2->2->1

Output: true

Follow up:

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

 

1.stack。O(n), O(n)

快慢指针找出中点，在找中点的过程中把前半段压入栈。接下来对比中点后半段的内容和stack pop出来的内容是否一致，出现不一致就是false，遍历完为true。

2.半段链表反转。O(n), O(1)。for follow up。

快慢指针找出中点，把中点后半段链表反转，接着从head和中点开始并行往前，一个个对比。 

 

实现1：

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
class Solution {
    public boolean isPalindrome(ListNode head) {
        Stack<Integer> stack = new Stack<>();
        ListNode fast = head, slow = head;
        // 1-2-3-4    32  null3
        // 1-2-3      32
        while (fast != null && fast.next != null) {
            fast = fast.next.next;
            stack.push(slow.val);
            slow = slow.next;
        }
        if (fast != null) {
            slow = slow.next;
        }
        while (!stack.isEmpty()) {
            if (stack.peek() != slow.val) {
                return false;
            }
            stack.pop();
            slow = slow.next;
        }
        return true;
    }
}

 

实现2：

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
class Solution {
    public boolean isPalindrome(ListNode head) {
        ListNode mid = findMiddle(head);
        mid = reverseList(mid);
        while (head != null && mid != null) {
            if (head.val != mid.val) {
                return false;
            }
            head = head.next;
            mid = mid.next;
        }
        return true;
    }
    
    private ListNode findMiddle(ListNode head) {
        ListNode fast = head, slow = head;
        // 1-2-3-4    32  null3
        // 1-2-3      32
        while (fast != null && fast.next != null) {
            fast = fast.next.next;
            slow = slow.next;
        }
        // return the first node after the midpoint. 1234 -> 3, 12345 -> 4
        if (fast != null) {
            slow = slow.next;
        }
        return slow;
    }
    
    private ListNode reverseList(ListNode head) {
        ListNode prev = null;
        while (head != null) {
            ListNode temp = head.next;
            head.next = prev;
            prev = head;
            head = temp;
        }
        return prev;
    }
}