Swap Nodes in Pairs 链表两两交换节点@LeetCode

题目：

链表两两交换节点

思路：

判断比较花时间，需要分奇数个节点和偶数个节点的情况，待改进。

package Level2;

import Utility.ListNode;

/**
 * Swap Nodes in Pairs 
 * 
 * 	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 S24 {

	public static void main(String[] args) {
		ListNode n1 = new ListNode(1);
		ListNode n2 = new ListNode(2);
		n1.next = n2;
		ListNode n3 = new ListNode(3);
		n2.next = n3;
		n1.print();
		ListNode head = swapPairs(n1);
		head.print();
	}
	
	public static ListNode swapPairs(ListNode head) {
        if(head == null){
        	return null;
        }
        // 当只有一个元素的情况
        if(head.next == null){
        	return head;
        }
        ListNode i = head;	// i指向第1个
        ListNode j = i.next;	// j指向第2个
        ListNode k = j.next;	// k指向第3个
        
        head = head.next;	
        while(j != null){
        	j.next = i;
        	if(k!=null && k.next!=null){	// 当有偶数个节点 
        		i.next = k.next;
        	}else{		// 当有奇数个节点
        		i.next = k;
        	}
        	
        	// 更新i，j，k的值，前进两格
        	i = k;
        	if(k != null){
        		j = k.next;
        	}else{
        		j = null;
        	}
        	if(k!=null && k.next!=null){
        		k = k.next.next;
        	}else{
        		k = null;
        	}
        }
        return head;
	}
}

重写：

用dummyHead解决奇偶问题，只要两个指针变量就够了

public ListNode swapPairs(ListNode head) {
        if(head == null || head.next == null){
        	return head;
        }
        
        ListNode dummyHead = new ListNode(0);
        dummyHead.next = head;
        ListNode cur = dummyHead;
        ListNode probe = cur.next;
        while(probe!=null && probe.next!=null){
            cur.next = probe.next;
            probe.next = probe.next.next;
            cur.next.next = probe;
            cur = probe;
            probe = probe.next;
        }
        return dummyHead.next;
    }

/**
 * 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) {
        if(head==null || head.next==null){
            return head;
        }
        ListNode dm = new ListNode(0);
        dm.next = head;
        ListNode p=dm, q=head, r;
        while(q!=null && q.next!=null){
            p.next = q.next;
            r = q.next.next;
            p.next.next = q;
            q.next = r;
            p = q;
            q = r;
        }
        return dm.next;
    }
}

递归法:

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