0328. 奇偶链表

• 链表
• 双指针
• 哑节点

328. 奇偶链表

这题双指针，配合哑节点，进行移动，将一个链表重新组装成两个链表，最后偶链表接在奇链表的后面。需要注意的是指针移动的条件，不满足偶数节点为空或偶数节点下节点为空 这种情况下才进行移动。
!(pOu == null || pOu.next == null)

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode() {}
 *     ListNode(int val) { this.val = val; }
 *     ListNode(int val, ListNode next) { this.val = val; this.next = next; }
 * }
 */
class Solution {
    public ListNode oddEvenList(ListNode head) {
        if (head == null || head.next == null) {
            return head;
        }
        
        ListNode pJi = head;
        ListNode pOu = head.next;

        ListNode dummy1 = new ListNode();
        ListNode dummy2 = new ListNode();
        ListNode tail1 =dummy1;
        ListNode tail2 = dummy2;

        while(true){
            
            tail1.next = pJi;
            tail2.next = pOu;
            tail1 = tail1.next;
            tail2 = tail2.next;

            if( !(pOu == null || pOu.next == null) ){
                pJi = pJi.next.next;
                pOu = pOu.next.next;
            }else{
                break;
            }
            tail1.next = null;
            tail2.next = null;
        }
        tail1.next = dummy2.next;
        return dummy1.next;

    }
}