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;
    }
}

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;
    }
}

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