24. Swap Nodes in Pairs

原创于 2016-04-17 10:56:32 发布 · 291 阅读

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本文详细介绍了如何通过递归方法解决链表中相邻节点互换的问题，给出了具体实现代码并解释了核心逻辑。

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.

public class Solution{
	public ListNode swapPairs(ListNode head) {
		if (head == null || head.next == null)
			return head;
		
		ListNode second = head.next;
		ListNode third = second.next;
		
		second.next = head;
		head.next = swapPairs(third);
		
		return second;
	}
}
//从题目的意思上看，就知道这题目有很多重复性的动作，可以通过递归来解决，不难~