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) {
        if(head == null){
            return head;
        }
        // write code here
        ListNode slow = head.next;
        ListNode node = head;
        ListNode fast = slow;
        while(fast != null && fast.next != null){
            node.next = fast.next;
            node = node.next;
            fast.next = node.next;
            fast = fast.next;
        }
        node.next = slow;
        
        return head;
    }
}