Palindrome Linked List 综合了反转链表和快慢指针的解法

这道题容易想到的思路是：遍历链表存入list中，判断list是否回文

但综合类解法为：用快慢指针找到链表的中间位置，然后将链表从中间断开，将后半部分翻转，与前半部分比较。

class Solution {
    public boolean isPalindrome(ListNode head) {
        List<Integer> list = new LinkedList<>();
        while(head != null){
            list.add(head.val);
            head = head.next;
        }
        int i = 0;
        int j = list.size()-1;
        while(i<j){
            if(!list.get(i).equals(list.get(j))) //list中两个元素相等判断方法equals
                return false;
            i++;
            j--;
        }
        return true;
    }
}

class Solution {
    public boolean isPalindrome(ListNode head) {
        if(head == null)return true;
        ListNode p1 = head;
        ListNode mid = slowfast(head);
        ListNode second = reverse(mid);
        ListNode p2 = second;
        boolean res = true;
        while(true && p2 != null){
            if(p1.val != p2.val) res = false;
            p1 = p1.next;
            p2 = p2.next;
        }
        mid.next = reverse(second);
        return res;
    }
    //快慢指针法
    public ListNode slowfast(ListNode head){
        ListNode slow = head;
        ListNode fast = head;
        while(fast != null && fast.next != null){
            slow = slow.next;
            fast = fast.next.next;
        }
        return slow;
    }
    //链表反转
    public ListNode reverse(ListNode head){
        ListNode pre = null;
        ListNode cur = head;
        while(cur != null){
            ListNode last = cur.next;
            cur.next = pre;
            pre = cur;
            cur = last;
        }
        return pre;
    }
}