package leetcode_1To300;

/**
 * 本代码来自 Cspiration，由 @Cspiration 提供
 * 题目来源：http://leetcode.com
 * - Cspiration 致力于在 CS 领域内帮助中国人找到工作，让更多海外国人受益
 * - 现有课程：Leetcode Java 版本视频讲解（1-900题）（上）（中）（下）三部
 * - 算法基础知识（上）（下）两部；题型技巧讲解（上）（下）两部
 * - 节省刷题时间，效率提高2-3倍，初学者轻松一天10题，入门者轻松一天20题
 * - 讲师：Edward Shi
 * - 官方网站：https://cspiration.com
 * - 版权所有，转发请注明出处
 */
public class _143_ReorderList {
    /**
     * 143. Reorder List
     * Given a singly linked list L: L0→L1→…→Ln-1→Ln,
     reorder it to: L0→Ln→L1→Ln-1→L2→Ln-2→…

     You must do this in-place without altering the nodes' values.

     For example,
     Given {1,2,3,4}, reorder it to {1,4,2,3}.

     time : O(n)
     space : O(1)

     * @param head
     */
    public void reorderList(ListNode head) {
        if (head == null || head.next == null) return;
        ListNode dummy = new ListNode(0);
        dummy.next = head;
        ListNode temp = null;
        ListNode slow = head, fast = head;
        ListNode l1 = head;
        while (fast != null && fast.next != null) {
            temp = slow;
            slow = slow.next;
            fast = fast.next.next;
        }
        temp.next = null;
        ListNode l2 = reverse(slow);
        merge(l1, l2);
    }

    private ListNode reverse(ListNode head) {
        ListNode prev = null;
        while (head != null) {
            ListNode temp = head.next;
            head.next = prev;
            prev = head;
            head = temp;
        }
        return prev;
    }

    private void merge(ListNode l1, ListNode l2) {
        while (l1 != l2) {
            ListNode n1 = l1.next;
            ListNode n2 = l2.next;
            l1.next = l2;
            if (n1 == null) break;
            l2.next = n1;
            l1 = n1;
            l2 = n2;
        }
    }
}