Reorder List

Reorder List

问题：

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.

思路：

　　使用HashMap记录位置，数学简易推导，进行List的变换

我的代码：

public class Solution {

    public void reorderList(ListNode head) {

        if(head == null)    return;

        int count = 0;

        HashMap<Integer,ListNode> hm = new HashMap<Integer,ListNode>();

        while(head != null)

        {

            hm.put(count++,head);

            head = head.next;

        }

        int sum = count - 1;

        for(int i = 0; i < count/2; i++)

        {

            ListNode cur = hm.get(i);

            ListNode next = hm.get(i + 1);

            ListNode last = hm.get(sum - i);

            cur.next = last;

            last.next = next;

            next.next = null;

        }

        return;

    }

}

View Code

他人代码：

public class Solution {



    private ListNode start;



    public void reorderList(ListNode head) {



        // 1. find the middle point

        if(head == null || head.next == null || head.next.next == null)return;



        ListNode a1 = head, a2 = head;



        while(a2.next!=null){

            // a1 step = 1

            a1 = a1.next;

            // a2 step = 2

            a2 = a2.next;

            if(a2.next==null)break;

            else a2 = a2.next;

        }

        // a1 now points to middle, a2 points to last elem



        // 2. reverse the second half of the list

        this.reverseList(a1);



        // 3. merge two lists

        ListNode p = head, t1 = head, t2 = head;

        while(a2!=a1){ // start from both side of the list. when a1, a2 meet, the merge finishes.

            t1 = p;

            t2 = a2;

            p = p.next;

            a2 = a2.next;



            t2.next = t1.next;

            t1.next = t2;

        }

    }



    // use recursion to reverse the right part of the list

    private ListNode reverseList(ListNode n){



        if(n.next == null){

            // mark the last node

            // this.start = n;

            return n;

        }



        reverseList(n.next).next = n;

        n.next = null;

        return n;

    }

}

View Code

学习之处：

• 他人的代码的思路更好，因为节省了空间，最后还是转换成了追击问题，通过追击问题，确定中间点进行分割，前面List和反转后的后面List进行Merge