[LeetCode]Swap Nodes in Pairs

很简单题，值得注意的是因为head会变，所以加一个sentinel节点，sentinel节点都能很优雅地解决head需要改变的问题。另外要注意的是更新curr的时候，由于在swap的时候，curr的为止变了，所以不能用curr = curr.next.next。

代码如下：

	/**
	* Definition for singly-linked list.
	* public class ListNode {
	* int val;
	* ListNode next;
	* ListNode(int x) {
	* val = x;
	* next = null;
	* }
	* }
	*/
	public class Solution {
	public ListNode swapPairs(ListNode head) {
	ListNode sentinel = new ListNode(0);
	sentinel.next = head;
	ListNode pre = sentinel;
	ListNode curr = head;
	while (curr != null && curr.next != null) {
	pre.next = swap(curr, curr.next);
	pre = pre.next.next;
	curr = pre.next;
	}
	return sentinel.next;
	}
	private ListNode swap(ListNode l1, ListNode l2) {
	l1.next = l2.next;
	l2.next = l1;
	return l2;
	}
	}

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