24. Swap Nodes in Pairs

题目：

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.

链接：https://leetcode.com/problems/swap-nodes-in-pairs/#/description

4/12/2017

4ms, 85%

链表题画画图比较好

public class Solution {
    public ListNode swapPairs(ListNode head) {
        if (head == null || head.next == null) return head;
        ListNode dummy = new ListNode(-1);
        dummy.next = head;
        ListNode prev = dummy;
        ListNode cur = head;

        while(cur != null && cur.next != null) {
            prev.next = cur.next;
            cur.next = cur.next.next;
            prev.next.next = cur;
            prev = cur;
            cur = cur.next;
        }
        return dummy.next;
    }
}

别人的recursive做法，虽然空间复杂度高，但是这种思路挺有意思，贴出来的代码比较容易懂一点

https://discuss.leetcode.com/topic/4351/my-accepted-java-code-used-recursion

https://discuss.leetcode.com/topic/41459/java-simple-recursive-solution

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;
    }
}

更多讨论：

https://discuss.leetcode.com/category/32/swap-nodes-in-pairs