2022.07.29_Daily Question

328. 奇偶链表

题目描述

给定单链表的头节点 head ,将所有索引为奇数的节点和索引为偶数的节点分别组合在一起,然后返回重新排序的列表.

第一个节点的索引被认为是 奇数 , 第二个节点的索引为 偶数 ,以此类推.

请注意,偶数组和奇数组内部的相对顺序应该与输入时保持一致.

你必须在 O(1) 的额外空间复杂度和 O(n) 的时间复杂度下解决这个问题.

 

示例 1:

输入: head = [1,2,3,4,5]
输出: [1,3,5,2,4]

示例 2:

输入: head = [2,1,3,5,6,4,7]
输出: [2,3,6,7,1,5,4]

 

提示:

• n ==  链表中的节点数
• 0 <= n <= 104
• -106 <= Node.val <= 106

coding

/** * Definition for singly-linked list. * public class ListNode { * int val; * ListNode next; * ListNode() {} * ListNode(int val) { this.val = val; } * ListNode(int val, ListNode next) { this.val = val; this.next = next; } * } */
class Solution {
    
    public ListNode oddEvenList(ListNode head) {
    
        if (head == null || head.next == null) {
    
            return head;
        }
        ListNode node1 = head;
        ListNode node2 = head.next;
        // Holds the node at the even index position
        ListNode temp = head.next;
        ListNode cur = head.next.next;
        // flag = true -> 奇数索引
        boolean flag = true;  
        while (cur != null) {
    
            if (flag) {
    
                node1.next = cur;
                node1 = node1.next;
            } else {
    
                node2.next = cur;
                node2 = node2.next;
            }
            cur = cur.next;
            flag = !flag;
        }
        // node1 -> node2
        // 如果 node1.next = head.next; 会出现 Error - Found cycle in the ListNode, One-way circular linked list appears (因为此时的headThe original linked list has been changed)
        node1.next = temp;
        node2.next = null;
        return head;
    }
}