题目要求
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| Given a linked list, swap every two adjacent nodes and return its head.
For example,
Given 1->2->3->4, you should return the list as 2->1->4->3.
Your algorithm should use only constant space. You may not modify the values in the list, only nodes itself can be changed.
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翻译过来就是：将链表中相邻两个节点交换顺序，并返回最终的头节点。
思路
这题的核心解题思路在于如何不占用额外的存储空间，就改变节点之间的关系。我们设置了一个节点pre，为当前一个节点的前一个节点
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| public ListNode swapPairs(ListNode head) {
ListNode start = new ListNode(0);
start.next = head;
ListNode pre = start;
while(head!=null && head.next!=null){
pre.next = head.next;
head.next = pre.next.next;
pre.next.next = head;
pre = pre.next.next;
head = pre.next;
}
return start.next;
}
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