LeetCode解题报告:Reorder List

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}.

思路：

1.利用快慢两个指针将链表一分为二；

2.针对第二个子链表求倒序；

3.利用merge思想将两个子链表合并。

代码：

public class ReorderSingleList {
    /**
     * Definition for singly-linked list. class ListNode { int val; ListNode
     * next; ListNode(int x) { val = x; next = null; } }
     */
    public void reorderList(ListNode head) {
        if (head == null || (head.next == null)) {
            return;
        }
        // fast and slow point find the mid position.
        ListNode fast = head;
        ListNode slow = head;
        while ((fast != null) && (fast.next != null)) {
            fast = fast.next.next;
            slow = slow.next;
        }

        // reverse the last second list.
        ListNode headnode = new ListNode(-1);
        headnode.next = slow;
        ListNode temp = headnode.next;
        while (temp.next != null) {
            ListNode insert = temp.next;
            ListNode currNext = insert.next;
            insert.next = headnode.next;
            headnode.next = insert;
            temp.next = currNext;
        }

        // merge insert
        ListNode firstcur = head;
        ListNode secondcur = headnode.next;
        while (firstcur != slow && (secondcur != slow)) {// at first,I make a mistake in here;
            ListNode firstnex = firstcur.next;
            ListNode secondnex = secondcur.next;
            firstcur.next = secondcur;
            secondcur.next = firstnex;
            firstcur = firstnex;
            secondcur = secondnex;
        }
    }

    public static void main(String[] args) {
        ListNode t5 = new ListNode(5);
        ListNode t4 = new ListNode(4, t5);
        ListNode t3 = new ListNode(3, t4);
        ListNode t2 = new ListNode(2, t3);
        ListNode t1 = new ListNode(1, t2);
        new ReorderSingleList().reorderList(t1);
        System.out.println(t1);
    }

}