题解 | #链表的奇偶重排#

import java.util.*;

/*
 * 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;
         ListNode fake_head1 = new ListNode(-1);
         ListNode fake_head2 = new ListNode(-1);
         ListNode pre1 = fake_head1,pre2 = fake_head2;
         int count = 1;
         for(ListNode cur = head;cur != null;cur = cur.next,count++){
             if((count&1) == 0){
                 pre2.next = new ListNode(cur.val);
                 pre2 = pre2.next;
             }else {
                 pre1.next = new ListNode(cur.val);
                 pre1 = pre1.next;
             }
         }
        pre1.next = fake_head2.next;
        return fake_head1.next;
    }
}


/** 
    看错了题,以为交换值的奇偶节点!
   // 可以认定为,将找到所有奇数节点,将节点插到前面!
        ListNode fake_head = new ListNode(-1);
        fake_head.next = head;
        ListNode  pre = fake_head,rush = head,rush_pre = fake_head;
        while(rush != null){
            while(rush != null && (rush.val&1) == 0) {
                rush_pre = rush_pre.next;
                rush = rush.next;
            }
            
            if(rush == null) break;
            if(pre.next == rush){
                rush = rush.next;
                pre = pre.next;
                rush_pre = rush_pre.next;
                continue;
            }
            // 断开旧链
            rush_pre.next = rush.next;
            
            // 插入新链
            rush.next = pre.next;
            pre.next = rush;
            
            // 继续向后探寻!
            rush = rush_pre.next;
            pre = pre.next;
        }
        return fake_head.next;
**/