题解 | #链表的奇偶重排#

public class Solution {

/**
用双指针
*/
public ListNode oddEvenList (ListNode head) {
    // write code here
    if(head==null) return null;
    ListNode dummy=new ListNode(-1);
    dummy.next=head;
    ListNode head1=head.next;
    ListNode p1=head;
    ListNode p2=head.next;
    while(p1.next!=null && p2.next!=null){
        p1.next=p2.next;
        p1=p1.next;
        p2.next=p1.next;
        p2=p2.next;
    }
    p1.next=head1;
    return dummy.next;
}

}