Parity rearrangement of Bm14 linked list

Ideas : This problem can be solved by skillfully using the correlation between parity

public class Solution {
    
    /** *  The class name in the code 、 Method name 、 The parameter name has been specified , Do not modify , Return the value specified by the method directly  * * @param head ListNode class  * @return ListNode class  */
    public ListNode oddEvenList (ListNode head) {
    
        // write code here
        if(head == null){
    
            return null;
        }
        // 1:  Record the first parity node 
        ListNode p1Head = head;
        ListNode p2Head = head.next;
        ListNode p1 = p1Head;
        ListNode p2 = p2Head;
        // 2:  Odd and even nodes use each other to connect their own nodes 
        // Be careful : The end condition should ensure that the odd node is not empty , Because it depends on the last odd node to connect the first even node 
        // Therefore, the end condition can be determined according to the odd number of nodes and the even number of nodes 
        while(p2 != null && p2.next != null){
    
            p1.next = p2.next;
            p1 = p1.next;
            p2.next = p1.next;
            p2 = p2.next;
        }
        p1.next = p2Head;
        return head;
    }
}