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}.

这道题是关于链表的综合性的题目，考察了对链表的多种操作。思路主要为：

1. 找到中间点，将链表分两半

2. 将第二半reverse

3. merge两个新链表

代码如下：

/**
 * Definition for singly-linked list.
 * class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) {
 *         val = x;
 *         next = null;
 *     }
 * }
 */
public class Solution {
    public void reorderList(ListNode head) {
        if(head == null || head.next == null){
            return;
        }
        ListNode mid = findMid(head);
        ListNode head2 = reverseList(mid.next);
        mid.next = null;
        mergeTwoLists(head, head2);
        
        
    }
    
    
    public ListNode findMid(ListNode head){
        ListNode fast = head.next;
        ListNode slow = head;
        while(fast != null && fast.next != null){
            slow = slow.next;
            fast = fast.next.next;
        }
        return slow;
    }
    
    public ListNode reverseList(ListNode head){
        ListNode nextNode = head.next;
        head.next = null;
        while(nextNode != null){
            ListNode temp = nextNode.next;
            nextNode.next = head;
            head = nextNode;
            nextNode = temp;
        }
        return head;
    }
    
    public ListNode mergeTwoLists(ListNode head1, ListNode head2){
        ListNode dummy = new ListNode(-1);
        ListNode tail = dummy;
        boolean isList1 = true;
        while(head1 != null && head2 != null){
            if(isList1){
                tail.next = head1;
                head1 = head1.next;
                isList1 = false;
            }
            else if(!isList1){
                tail.next = head2;
                head2 = head2.next;
                isList1 = true;
            }
            tail=tail.next;
        }
        while(head1 != null){
            tail.next = head1;
            head1 = head1.next;
            tail =tail.next;
        }
        
        while(head2 != null){
            tail.next = head2;
            head2 = head2.next;
            tail = tail.next;
        }
        
        return dummy.next;
        
    }
    
    
}