234. 回文链表 - 力扣（LeetCode）

1.转化法

思路

        将链表转化为列表进行比较

复习到的知识

        arraylist的长度函数：list.size()

具体代码

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode() {}
 *     ListNode(int val) { this.val = val; }
 *     ListNode(int val, ListNode next) { this.val = val; this.next = next; }
 * }
 */
class Solution {
    public boolean isPalindrome(ListNode head) {
        ListNode n = head;
        List<Integer> list = new ArrayList<>();
        while(n!=null){
            list.add(n.val);
            n=n.next;
        }
        int l=0;
        int r=list.size()-1;
        while(l<r){
            if(list.get(l)!=list.get(r)){
                return false;
            }
            l++;
            r--;
        }
        return true;

    

    }
}

 2.反转法

思路

        将后半段链表反转，然后进行比较。

知识

        取单链表的中间节点：快慢指针

具体代码

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode() {}
 *     ListNode(int val) { this.val = val; }
 *     ListNode(int val, ListNode next) { this.val = val; this.next = next; }
 * }
 */
class Solution {
    public boolean isPalindrome(ListNode head) {
        ListNode n1 = secondL(head);
        ListNode n2 = reverseL(n1.next);
        ListNode p1 = head;
        ListNode p2 = n2;
        


        while(p2!=null){
            if(p1.val != p2.val){
                return false;
            }

            p1=p1.next;
            p2=p2.next;
        }
        return true;
        
    }

    public ListNode reverseL(ListNode l){
        ListNode old = null;
        ListNode current = l;
        while(current!=null){
            ListNode tem = current.next;
            current.next = old;
            old =current;
            current= tem;
        }
        return old;
    }

    public ListNode secondL(ListNode l){
        ListNode slow = l;
        ListNode fast = l;
        while(fast.next!=null && fast.next.next!=null){
            slow = slow.next;
            fast = fast.next.next;
        }
        return slow;
    }
}