题解 | #链表的奇偶重排#

function oddEvenList( head ) {
    if (head === null || head.next === null);
    
    let dummyOdd = new ListNode(0);
    let dummyEven = new ListNode(0); 
    let p1 = dummyOdd;
    let p2 = dummyEven;
    
    let count = 1; 
    let curr = head;
    
    while (curr) {
        if (count % 2 === 0) {
            p2.next = curr;
            p2 = p2.next;
        } else {
            p1.next = curr;
            p1 = p1.next;
        }
        curr = curr.next;
        count++;
    }
    p2.next = null;
    p1.next = dummyEven.next;

    return dummyOdd.next;
}