题解 | #链表的奇偶重排#

/*
 * public class ListNode {
 *   int val;
 *   ListNode next = null;
 *   public ListNode(int val) {
 *     this.val = val;
 *   }
 * }
 */

public class Solution {
    /**
     * 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
     *
     * 
     * @param head ListNode类 
     * @return ListNode类
     */
    public ListNode oddEvenList (ListNode head) {
        // write code here
        if(head == null || head.next == null || head.next.next == null){
            return head;
        }
        ListNode result = new ListNode(-1);
        
        ListNode currentNode = result;
        ListNode p1 = head;
        ListNode p2 = head;
        int count = 1;
        while(p1 != null){
            if( (count & 1 ) == 1){
                currentNode.next = new ListNode(p1.val);
                currentNode = currentNode.next;
            }
            p1 = p1.next;
            count ++;
        }
        
        count = 1;
        while(p2 != null){
            if( (count & 1 ) == 0){
                currentNode.next = new ListNode(p2.val);
                currentNode = currentNode.next;
            }
            p2 = p2.next;
            count ++;
        }
        
        
        return result.next;
        
    }
}