题目:回文链表定义为链表正序和逆序输出结果一致，检验链表是否是回文链表.

解法一

可以将链表反转之后和现在的链表的顺序输出做对比.
<pre><code>` func isPalindrome(node:ListNode) -> Bool {

    var headNode:ListNode? = node
    var preNode:ListNode?
    var reverseHeadNode:ListNode?
    
    while headNode != nil {
        let next:ListNode? = headNode?.next
        
        if next == nil {
            reverseHeadNode = headNode
        }
        
        headNode?.next = preNode
        preNode = headNode
        
        headNode = next
    }
    
    headNode = node
    while headNode != nil {
        if  headNode?.value != reverseHeadNode?.value {
            return false
        }
        headNode = headNode?.next
        reverseHeadNode = reverseHeadNode?.next
    }
    return true
}`</code></pre>

解法二

可以先将链表遍历到链表中间，用数组将数据存储起来，然后继续遍历链表与数组最后的数字做对比.
<pre><code>` func isPalindrome1(node:ListNode) -> Bool {

    var list:[String?] = []
    
    var slow:ListNode? = node
    var fast:ListNode? = node
    
    while fast != nil && fast?.next != nil {
        list.append(slow?.value)
        slow = slow?.next
        fast = fast?.next?.next
    }
    
    // 奇数个元素，跳过中间元素
    if fast != nil {
        slow = slow?.next
    }
    
    while slow != nil {
        let data:String? = list.last!
        list.removeLast()
        if  slow?.value != data {
            return false
        }
        slow = slow?.next
    }
    
    return true
}`</code></pre>

测试代码:
<pre><code>`listNodeManger.headNode = nil

listNodeManger.appendToTail(value: "(1)")
listNodeManger.appendToTail(value: "(2)")
listNodeManger.appendToTail(value: "(3)")
listNodeManger.appendToTail(value: "(2)")
listNodeManger.appendToTail(value: "(1)")

var isRome:Bool = listNodeManger.isPalindrome(node: listNodeManger.headNode!)
if isRome {
print("FlyElephant---回文链表")
} else {
print("FlyElephant---非回文链表")
}

isRome = listNodeManger.isPalindrome1(node: listNodeManger.headNode!)
if isRome {
print("FlyElephant---回文链表")
} else {
print("FlyElephant---非回文链表")
}
`</code></pre>

FlyElephant.png