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;    }}