数据结构与算法(Java) 28:判断一个链表是否为回文结构

题目 给定一个链表的头节点head,请判断该链表是否为回 文结构。 例如: 1->2->1,返回true。 1->2->2->1,返回true。 15->6->15,返回true。 1->2->3,返回false。

进阶 如果链表长度为N,时间复杂度达到O(N),额外空间复杂 度达到O(1)。

思路 (1)额外空间复杂度O(N):将链表的值依次压栈,然后依次弹出并与原链表逐个比较;

        (2)额外空间复杂度O(N/2):设置两个指针p,q,其中p一次走一步,q一次走两步,待q遍历到最后p便位于链表中间,然后将链表后半部分一次压栈,然后依次弹出与前半部分链表进行比较;

        (3)额外空间复杂度O(1):设置两个指针p,q,其中p一次走一步,q一次走两步,待q遍历到最后p便位于链表中间,然后将链表后半部分数值进行反转【也就是反转单向链表结构操作】,从两头开始遍历并逐个比较。

package algorithm.section4;

import java.util.Stack;

public class IsPalindromeList {
    public static class Node{
        int value;
        Node next;

        public Node(int value){
            this.value = value;
        }
    }
    public static boolean method1(Node head){
        if (head == null || head.next == null) return true;
        Stack<Integer> flag = new Stack<Integer>();
        Node p = head;
        while (p != null){
            flag.push(p.value);
            p = p.next;
        }
        while (!flag.isEmpty()){
            if (head.value != flag.pop())
                return false;
            head = head.next;
        }
        return true;
    }
    public static boolean method2(Node head){
        if (head == null || head.next == null) return true;
        Node p = head;
        Node q = head;
        while (q.next != null && q.next.next != null){
            p = p.next;
            q = q.next.next;
        }
        Stack<Integer> flag = new Stack<Integer>();
        Node p0 = p;
        while (p.next != null){
            flag.push(p.next.value);
            p = p.next;
        }
        while (!flag.isEmpty()){
            if (head.value != flag.pop())
                return false;
            head = head.next;
        }
        return true;
    }
    public static boolean method3(Node head){
        if (head == null || head.next == null) return true;
        Node p = head;
        Node q = head;
        while (q.next != null && q.next.next != null){
            p = p.next;
            q = q.next.next;
        }
        Node p0 = p.next;
        p.next = null;
        while (p0 != null){
            q = p0.next;
            p0.next = p;
            p =  p0;
            p0 = q;
        }
        p0 = p;
        p = head;
        while (p != null && p0 != null){
            if (p.value != p0.value)
                return false;
            p = p.next;
            p0 = p0.next;
        }
        return true;
    }

    public static void main(String[] args){
        Node head = null;
        System.out.print(method1(head) + " | ");
        System.out.print(method2(head) + " | ");
        System.out.println(method3(head) + " | ");
        System.out.println("=========================");

        head = new Node(1);
        System.out.print(method1(head) + " | ");
        System.out.print(method2(head) + " | ");
        System.out.println(method3(head) + " | ");
        System.out.println("=========================");

        head = new Node(1);
        head.next = new Node(2);
        System.out.print(method1(head) + " | ");
        System.out.print(method2(head) + " | ");
        System.out.println(method3(head) + " | ");
        System.out.println("=========================");

        head = new Node(1);
        head.next = new Node(1);
        System.out.print(method1(head) + " | ");
        System.out.print(method2(head) + " | ");
        System.out.println(method3(head) + " | ");
        System.out.println("=========================");

        head = new Node(1);
        head.next = new Node(2);
        head.next.next = new Node(3);
        System.out.print(method1(head) + " | ");
        System.out.print(method2(head) + " | ");
        System.out.println(method3(head) + " | ");
        System.out.println("=========================");

        head = new Node(1);
        head.next = new Node(2);
        head.next.next = new Node(1);
        System.out.print(method1(head) + " | ");
        System.out.print(method2(head) + " | ");
        System.out.println(method3(head) + " | ");
        System.out.println("=========================");

        head = new Node(1);
        head.next = new Node(2);
        head.next.next = new Node(3);
        head.next.next.next = new Node(1);
        System.out.print(method1(head) + " | ");
        System.out.print(method2(head) + " | ");
        System.out.println(method3(head) + " | ");
        System.out.println("=========================");

        head = new Node(1);
        head.next = new Node(2);
        head.next.next = new Node(2);
        head.next.next.next = new Node(1);
        System.out.print(method1(head) + " | ");
        System.out.print(method2(head) + " | ");
        System.out.println(method3(head) + " | ");
        System.out.println("=========================");

        head = new Node(1);
        head.next = new Node(2);
        head.next.next = new Node(3);
        head.next.next.next = new Node(2);
        head.next.next.next.next = new Node(1);
        System.out.print(method1(head) + " | ");
        System.out.print(method2(head) + " | ");
        System.out.println(method3(head) + " | ");
        System.out.println("=========================");
    }
}
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转载自blog.csdn.net/Dorothy_Xue/article/details/105720888