题目

给定一个单链表 L 的头节点 head ，单链表 L 表示为：

L0 → L1 → … → Ln - 1 → Ln

请将其重新排列后变为：

L0 → Ln → L1 → Ln - 1 → L2 → Ln - 2 → …

不能只是单纯的改变节点内部的值，而是需要实际的进行节点交换。

示例 1：

输入：head = [1,2,3,4]
输出：[1,4,2,3]

示例 2：

输入：head = [1,2,3,4,5]
输出：[1,5,2,4,3]

提示：

链表的长度范围为 [1, 5 * 104]
1 <= node.val <= 1000

解答

源代码

/**
 * 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 void reorderList(ListNode head) {
        if (head.next == null) {
            return;
        }

        ListNode middleList = searchMiddleList(head);
        ListNode reversedList = reverse(middleList);
        ListNode res = linkList(head, reversedList);
    }

    // 寻找中间节点
    public ListNode searchMiddleList(ListNode head) {
        ListNode pre = new ListNode(0);
        ListNode fast = head, slow = head;
        pre.next = slow;

        while (fast != null && fast.next != null) {
            fast = fast.next.next;
            slow = slow.next;
            pre = pre.next;
        }

        // 将前半部分链表结尾置空
        pre.next = null;

        return slow;
    }

    // 反转链表
    public ListNode reverse(ListNode head) {
        ListNode pre = null;
        ListNode cur = head;

        while (cur != null) {
            ListNode next = cur.next;
            cur.next = pre;
            pre = cur;
            cur = next;
        }

        return pre;
    }

    // 交替拼接两个链表
    public ListNode linkList(ListNode head1, ListNode head2) {
        ListNode dummy = new ListNode(0);
        ListNode temp = dummy;

        while (head1 != null && head2 != null) {
            temp.next = head1;
            head1 = head1.next;
            temp = temp.next;
            temp.next = head2;
            head2 = head2.next;
            temp = temp.next;
        }

        if (head1 == null) {    // head1已经拼接完毕
            temp.next = head2;
        } else {                // head2已经拼接完毕
            temp.next = head1;
        }

        return dummy.next;
    }
}