【Leetcode】143. Reorder List

Question：

Given a singly linked list L: L₀→L₁→…→L_n-1→L_n,
reorder it to: L₀→L_n→L₁→L_n-1→L₂→L_n-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}.

Tips：

给定一个单链表，将链表重新排序，注意不能改变结点的值。

排序规则如下：

L₀→L₁→…→L_n-1→L_n,
L₀→L_n→L₁→L_n-1→L₂→L_n-2→…

思路：

重新排序后的链表，前1/2 结点相对顺序不变，而后半部分是逆序。所以我的思路是先将后半部分结点翻转，变为逆序，再将后半部分结点依次插入到前半部分中去。

大致分为三部分:

（1）找到链表的中间位置，将链表分为两部分。

（2）将第二部分链表逆序

（3）将第二部分所有节点依次插入到前半部分结点之间。

代码：

public void reorderList(ListNode head) {
        if (head == null || head.next == null)
            return;
        // Find the part2;第二部分是从slow.next开始的
        ListNode slow = head;
        ListNode fast = head;
        while (fast.next != null && fast.next.next != null) {
            slow = slow.next;
            fast = fast.next.next;
        }
        System.out.println("slow"+slow.val);
        ListNode mid = slow.next;
        slow.next = null;
        System.out.println("mid"+mid.val);
        // 将第二部分翻转;
        ListNode pre = null;
        ListNode cur = mid;
        while (cur != null) {
            if (cur.next != null) {
                ListNode next = cur.next;
                System.out.println("next"+next.val);
                cur.next = pre;
                pre = cur;
                cur = next;
            } else {
                cur.next = pre;
                pre = cur;
                cur=null;
            }
        }
        System.out.println("pre"+pre.val);
        // append one by one;
        ListNode p1 = head;
        ListNode p2 = pre;
        while (p2 != null) {
            ListNode n1 = p1.next;
            ListNode n2 = p2.next;
            p1.next = p2;
            p2.next = n1;
            p1 = p1.next.next;
            p1 = n1;
            p2 = n2;
        }    
        //print
        while (head != null) {
            System.out.println(head.val);
            head = head.next;
        }
    }

代码中的一些输出 是为了验证结果的正确性 提交时可删除。leetcode提交版版代码如下：

public void reorderList(ListNode head) {
        if (head == null || head.next == null)
            return;
        // Find the part2;第二部分是从slow.next开始的
        ListNode slow = head;
        ListNode fast = head;
        while (fast.next != null && fast.next.next != null) {
            slow = slow.next;
            fast = fast.next.next;
        }
        ListNode mid = slow.next;
        slow.next = null;
        // 将第二部分翻转;
        ListNode pre = null;
        ListNode cur = mid;
        while (cur != null) {
            if (cur.next != null) {
                ListNode next = cur.next;
                cur.next = pre;
                pre = cur;
                cur = next;
            } else {
                cur.next = pre;
                pre = cur;
                cur=null;
            }
        }
        // append one by one;
        ListNode p1 = head;
        ListNode p2 = pre;
        while (p2 != null) {
            ListNode n1 = p1.next;
            ListNode n2 = p2.next;
            p1.next = p2;
            p2.next = n1;
            p1 = p1.next.next;
            p1 = n1;
            p2 = n2;
        }
    }