143. Reorder List

Description

Given a singly linked list L: L₀→L₁→…→L_n-1→L_n,
reorder it to: L₀→L_n→L₁→L_n-1→L₂→L_n-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}.

Solution

Two-pointers

蛮简单的题，是几道题目的综合，思路如下：

• 将链表平均分成两半（快慢指针）；
• reverse后一个链表；
• merge两个链表。

/**
 * 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 || head.next.next == null) {
            return;
        }
        
        ListNode slow = head;
        ListNode fast = head.next;
        while (fast != null && fast.next != null) {
            slow = slow.next;
            fast = fast.next.next;
        }
        
        ListNode secondHead = slow.next;
        slow.next = null;
        secondHead = reverse(secondHead);
        mergeList(head, secondHead);
    }
    
    public ListNode reverse(ListNode head) {
        ListNode prev = null;
        while (head != null) {
            ListNode next = head.next;
            head.next = prev;
            prev = head;
            head = next;
        }
        return prev;
    }
    
    public ListNode mergeList(ListNode p, ListNode q) {
        ListNode dummy = new ListNode(0);
        ListNode tail = dummy;
        
        while (q != null) {
            tail.next = p;
            p = p.next;
            tail.next.next = q;
            q = q.next;
            tail = tail.next.next;
        }
        
        tail.next = p;
        return dummy.next;
    }
}