Monday, March 19, 2018

[2018-Interview] Palindrome Linked List

Original question: https://leetcode.com/problems/palindrome-linked-list/description/

Answer: Basic logic, find the middle node, and reverse the right-side nodes, then compare left-side nodes with right-side nodes. Answer is from Jiuzhang: https://www.jiuzhang.com/solution/palindrome-linked-list/



/**
* 本参考程序来自九章算法,由 @九章算法 提供。版权所有,转发请注明出处。
* - 九章算法致力于帮助更多中国人找到好的工作,教师团队均来自硅谷和国内的一线大公司在职工程师。
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* - 更多详情请见官方网站:http://www.jiuzhang.com/?source=code
*/ 

// This code would destroy the original structure of the linked list.
// If you do not want to destroy the structure, you can reserve the second part back.
public class Solution {
    /**
     * @param head a ListNode
     * @return a boolean
     */
    public boolean isPalindrome(ListNode head) {
        if (head == null) {
            return true;
        }
        
        ListNode middle = findMiddle(head);
        middle.next = reverse(middle.next);
        
        ListNode p1 = head, p2 = middle.next;
        while (p1 != null && p2 != null && p1.val == p2.val) {
            p1 = p1.next;
            p2 = p2.next;
        }
        
        return p2 == null;
    }
    
    private ListNode findMiddle(ListNode head) {
        if (head == null) {
            return null;
        }
        ListNode slow = head, fast = head.next;
        while (fast != null && fast.next != null) {
            slow = slow.next;
            fast = fast.next.next;
        }
        
        return slow;
    }
    
    private ListNode reverse(ListNode head) {
        ListNode prev = null;
        
        while (head != null) {
            ListNode temp = head.next;
            head.next = prev;
            prev = head;
            head = temp;
        }
        
        return prev;
    }
}

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