回文链表

请判断一个链表是否为回文链表。

示例 1:

输入: 1->2
输出: false

示例 2:

输入: 1->2->2->1
输出: true

参考答案

方案一

利用链表的后续遍历，使用函数调用栈作为后序遍历栈，来判断是否回文

var isPalindrome = function(head) {
    let left = head;
    function traverse(right) {
        if (right == null) return true;
        let res = traverse(right.next);
        res = res && (right.val === left.val);
        left = left.next;
        return res;
    }
    return traverse(head);
};

方案二

通过快、慢指针找链表中点，然后反转链表，比较两个链表两侧是否相等，来判断是否是回文链表

var isPalindrome = function(head) {
    // 反转 slower 链表
    let right = reverse(findCenter(head));
    let left = head;
    // 开始比较
    while (right != null) {
        if (left.val !== right.val) {
            return false;
        }
        left = left.next;
        right = right.next;
    }
    return true;
}
function findCenter(head) {
    let slower = head, faster = head;
    while (faster && faster.next != null) {
        slower = slower.next;
        faster = faster.next.next;
    }
    // 如果 faster 不等于 null，说明是奇数个，slower 再移动一格
    if (faster != null) {
        slower = slower.next;
    }
    return slower;
}
function reverse(head) {
    let prev = null, cur = head, nxt = head;
    while (cur != null) {
        nxt = cur.next;
        cur.next = prev;
        prev = cur;
        cur = nxt;
    }
    return prev;
}