24. Swap Nodes in Pairs

难度：Medium

题目要求

Given a linked list, swap every two adjacent nodes and return its head.

You may not modify the values in the list's nodes, only nodes itself may be changed.

Example:

Given 1->2->3->4, you should return the list as 2->1->4->3.

中文说明

给定一个链表，两两交换其中相邻的节点，并返回交换后的链表。

你不能只是单纯的改变节点内部的值，而是需要实际的进行节点交换。

思路1

循环迭代法。

public class Solution {
    public ListNode swapPairs(ListNode head) {
        if (head == null || head.next == null) return head;
        ListNode newHead = new ListNode(0);
        ListNode pprev = newHead;
        ListNode prev = head;
        ListNode cur = prev.next;
        while (cur != null) {
            prev.next = cur.next;
            cur.next = prev;
            pprev.next = cur;
            pprev = prev;
            prev = prev.next;
            if (prev != null)
                cur = prev.next;
            else
                cur = null;
        }
        return newHead.next;
    }
}

或者

public class Solution {
    public ListNode swapPairs(ListNode head) {
        ListNode dummy = new ListNode(0);
        dummy.next = head;
        ListNode current = dummy;
        while (current.next != null && current.next.next != null) {
            ListNode first = current.next;
            ListNode second = current.next.next;
            first.next = second.next;
            current.next = second;
            current.next.next = first;
            current = current.next.next;
        }
        return dummy.next;
    }
}

思路2

递归法。代码比迭代的方式优雅很多。

public class Solution {
    public ListNode swapPairs(ListNode head) {
        if (head == null || head.next == null) return head;
        ListNode next = head.next;
        head.next = swapPairs(next.next);
        next.next = head;
        return next;
    }
}

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