LeetCode234. 回文链表

时间O(n)空间O(1)的方法

 

var isPalindrome = function(head) {
    let left = head;
    let right;
    let slow=head ;let fast = head;
    while(fast !==null && fast.next !== null) {
　　　　// 通过快慢指针找中点
        fast = fast.next.next;
        slow = slow.next;
    }
    if(fast!==null) slow=slow.next; // 如果fast不等于null，说明链表是偶数，则slow需要再向下走一步
    let reverse = function(h) { //迭代反转链表的方法
        let nxt,cur = h;
        let pre = new ListNode(null);
        while(cur) {
            nxt = cur.next;
            cur.next = pre;
            pre = cur;
            cur = nxt;
        }
        return pre;
    }
    right = reverse(slow); // right等于反转后的slow
    while(right.next) { //如果right遍历完都相等的话就结束，因为left可能要比right的节点数多哦
        if(left.val !== right.val) {// 让left的val和right的val依次进行比较，如果不相等则说明不是回文
            return false;
        }
        left = left.next;
        right = right.next;
    }
    return true;
};