24. 两两交换链表中的节点

题意：

图示:

迭代图示(1)：

递归图示(2)：

解题思路：

思路：
    1. 添加虚拟头结点 dummy;
    2. 定义cur指向dummy；
    3. 初始化 $node1 = $cur->next; $node2 = $node1->next;
    4. 修改cur,node1,node2三者指向关系，如上图所示
    5. 移动cur结点到下一个结点，重复上面步骤，直到条件不满足为止
    6. 最后返回dummy->next

PHP代码实现：

/**
 * Definition for a singly-linked list.
 * class ListNode {
 *     public $val = 0;
 *     public $next = null;
 *     function __construct($val) { $this->val = $val; }
 * }
 */
class Solution {
    /**
     * @param ListNode $head
     * @return ListNode
     */
    function swapPairs($head) {
        $dummy = new ListNode(0);
        $dummy->next = $head;
        $cur = $dummy;
        while ($cur->next && $cur->next->next) {
            $node1 = $cur->next;
            $node2 = $node1->next;
            $next = $node2->next;
            $node2->next = $node1;
            $node1->next = $next;
            $cur->next = $node2;
            $cur = $node1;
        }
        return $dummy->next;
    }
    function swapPairsRecursive($head) {
        if ($head == null || $head->next == null) return $head;
        $next = $head->next;
        $head->next = $this->swapPairsRecursive($next->next);
        $next->next = $head;
        return $next;
    }
}

GO代码实现：

/**
 * Definition for singly-linked list.
 * type ListNode struct {
 *     Val int
 *     Next *ListNode
 * }
 */
func swapPairs(head *ListNode) *ListNode {
    if head == nil || head.Next == nil {
        return head
    }
    var dummy = &ListNode{}
    dummy.Next = head
    var cur = dummy
    for cur.Next != nil && cur.Next.Next != nil {
        node1 := cur.Next
        node2 := node1.Next
        next := node2.Next
        node2.Next = node1
        node1.Next = next
        cur.Next = node2
        cur = node1
    }
    return dummy.Next
}