leetcode -- 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}.

[解题思路]

将队列从中间断开，分成两个队列p1,p2，p2反转，然后将p1,p2进行合并

public class Solution {
    public void reorderList(ListNode head) {
        // IMPORTANT: Please reset any member data you declared, as
        // the same Solution instance will be reused for each test case.
        if(head == null){
            return;
        }
        
        ListNode fast = head, slow = head;
        while(fast.next != null && fast.next.next != null){
            fast = fast.next.next;
            slow = slow.next;
        }
        
        ListNode p1 = head, p2 = slow.next;
        // break from middle node
        slow.next = null;
        
        p2 = reverseList(p2);
        
        mergeTwoList(head, p1, p2);
    }
    
    public void mergeTwoList(ListNode head, ListNode p1, ListNode p2){
        if(p2 == null){
            return;
        }
        
        head = new ListNode(-65536);
        ListNode pResult = head;
        int count = 0;
        while(p1 != null && p2 != null){
            if(count % 2 == 0){
                pResult.next = p1;
                pResult = pResult.next;
                p1 = p1.next;
            } else {
                pResult.next = p2;
                pResult = pResult.next;
                p2 = p2.next;
            }
            count ++;
        }
        
        while(p1 != null){
            pResult.next = p1;
            pResult = pResult.next;
            p1 = p1.next;
        }
        while(p2 != null){
            pResult.next = p2;
            pResult = pResult.next;
            p2 = p2.next;
        }
        head = head.next;
    }
    
    public ListNode reverseList(ListNode head){
        if(head == null || head.next == null){
            return head;
        }
        ListNode cur = head.next, pre = head;
        ListNode tmp = null;
        pre.next = null;
        
        while(cur != null){
            tmp = cur.next;
            cur.next = pre;
            pre = cur;
            cur = tmp;
        }
        return pre;
    }
}