LeetCode 143. Reorder List

原题链接在这里：https://leetcode.com/problems/reorder-list/

题目：

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}.

题解：

找到中点，断开. 反转后半段，重新merge.

Time  Complexity: O(n). Space O(1).

AC Java:

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
class Solution {
    public void reorderList(ListNode head) {
        if(head == null || head.next == null){
            return;
        }
        
        ListNode mid = findMid(head);
        ListNode midNext = mid.next;
        mid.next = null;
        ListNode reverseMidNext = reverse(midNext);
        
        ListNode p = head;
        ListNode q = reverseMidNext;
        while(q!=null){
            ListNode qNext = q.next;
            q.next = p.next;
            p.next = q;
            
            q = qNext;
            p = p.next.next;
        }
        
        return;
    }
    
    private ListNode findMid(ListNode head){
        if(head == null || head.next == null){
            return head;
        }
        
        ListNode runner = head;
        ListNode walker = head;
        while(runner != null && runner.next != null && runner.next.next != null){
            walker = walker.next;
            runner = runner.next.next;
        }
        
        return walker;
    }
    
    private ListNode reverse(ListNode head){
        if(head == null || head.next == null){
            return head;
        }
        
        ListNode tail = head;
        ListNode cur = tail;
        ListNode pre;
        ListNode temp;
        while(tail.next != null){
            pre = cur;
            cur = tail.next;
            temp = cur.next;
            cur.next = pre;
            tail.next = temp;
        }
        
        return cur;
    }
}