234. [链表]回文链表

234. 回文链表

方法一：额外空间＋一次遍历

class Solution {
    public boolean isPalindrome(ListNode head) {
        List<Integer> list = new ArrayList<>();
        while (head != null) {
            list.add(head.val);
            head = head.next;
        }
        int front = 0, back = list.size() - 1;
        while(front < back){
            if(!list.get(front).equals(list.get(back))){
                return false;
            }
            front++;
            back--;
        }
        return true;
    }
}

方法二：反转中后段链表+双指针遍历

class Solution {
    public boolean isPalindrome(ListNode head) {
        if (head == null) {
            return true;
        }
        ListNode center = findCenterOfListNode(head);
        ListNode reverseCenter = reverseList(center.next);

        ListNode p1 = head;
        ListNode p2 = reverseCenter;
        while (p2 != null){
            if (p1.val != p2.val) return false;
            p1 = p1.next;
            p2 = p2.next;
        }
        return true;
    }
    
    private ListNode reverseList(ListNode head){
        ListNode prev = null;
        ListNode curr = head;
        while (curr != null){
            ListNode nextTemp = curr.next;
            curr.next = prev;
            prev = curr;
            curr = nextTemp;
        }
        return prev;
    }

    private ListNode findCenterOfListNode(ListNode head){
        ListNode fast = head;
        ListNode slow = head;
        while (fast.next != null && fast.next.next != null){
            fast = fast.next.next;
            slow = slow.next;
        }
        return slow;
    }
}