题解 | #链表的奇偶重排#

双指针

/*
 * 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){
            return  head;
        }
        ListNode oddP = head;
        ListNode evenNode = head.next;
        ListNode evenP = evenNode;
        while (oddP.next != null && evenP.next !=null){
            oddP.next = oddP.next.next;
            oddP = oddP.next;
            evenP.next = evenP.next.next;
            evenP = evenP.next;
        }
        oddP.next = evenNode;
        return  head;
    }
}