[Programmer's Guide interview Code] list issue - to determine whether a list is a palindrome (list)

topic

If that

answer

step

1 to find the midpoint of the list,
two flip right half,
3 disconnect sides, two from the same node list head start determination value.
4 and then flipped back right list

the complexity

Time O (n) space O (1)

Code

//public class Node{
//  int val;
//  Node next;
//  Node(int val){
//      this.val=val;
//  }
//}
public class Main {
    public static void main(String[] args) {
        Node n1=new Node(1);
        Node n2=new Node(2);
        Node n3=new Node(3);
        Node n4=new Node(3);
        Node n5=new Node(2);
        Node n6=new Node(1);
        
        n1.next=n2;
        n2.next=n3;
        n3.next=n4;
        n4.next=n5;
        n5.next=n6;
        Node head=n1;
        
        boolean ans=palindromeTree(head);
        System.out.println(ans);
    }
    
    public static boolean palindromeTree(Node head){
        //找到中点
        Node mid=head;//中点
        Node fast=head;
        while(fast.next!=null&&fast.next.next!=null) {//奇偶长度情况
            fast=fast.next.next;
            mid=mid.next;
        }
        
        //翻转右侧链表
        Node cur=mid.next;//
        mid.next=null;
        Node behind=null;
        Node before=null;
        while(cur!=null) {
            behind=cur.next;
            cur.next=before;
            before=cur;
            cur=behind;
        }
        Node rHead=before;
        
        //检查是否是回文的
        Node lNode=head;
        Node rNode=before;//
        boolean res=true;
        while(lNode!=null&&rNode!=null) {//原链奇偶长度都ok
            if(lNode.val!=rNode.val) {
                res=false;
                break;
            }
            lNode=lNode.next;
            rNode=rNode.next;
        }
        
        //右侧链表翻转回去
        before=rHead;
        cur=rHead.next;
        behind=null;
        while(cur!=null) {
            behind=cur.next;
            cur.next=before;
            before=cur;
            cur=behind;
        }
        
        return res;
    }
}