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Algorithm

143. Reorder List

Description

You are given the head of a singly linked-list. The list can be represented as:

L0 → L1 → … → Ln - 1 → Ln

Reorder the list to be on the following form:

L0 → Ln → L1 → Ln - 1 → L2 → Ln - 2 → …

You may not modify the values in the list's nodes. Only nodes themselves may be changed.

Example 1:

Input: head = [1,2,3,4]
Output: [1,4,2,3]

Example 2:

Input: head = [1,2,3,4,5]
Output: [1,5,2,4,3]

Constraints:

  • The number of nodes in the list is in the range [1, 5 * 104].
  • 1 <= Node.val <= 1000

Solution

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode() {}
 *     ListNode(int val) { this.val = val; }
 *     ListNode(int val, ListNode next) { this.val = val; this.next = next; }
 * }
 */
class Solution {
   public void reorderList(ListNode head) {
       if(head==null||head.next==null){
           return;
       }

       // Find the middle node
       ListNode slow = head, fast = head.next;
       while (fast != null && fast.next != null) {
           slow = slow.next;
           fast = fast.next.next;
       }

       // Reverse the second half
       ListNode head2 = reverse(slow.next);
       slow.next = null;

       // Intertwine the two halves
       merge(head, head2);
   }

   private void merge(ListNode head1, ListNode head2) {
       while (head2 != null) {
           ListNode next = head1.next;
           head1.next = head2;
           head1 = head2;
           head2 = next;
       }
   }

   private ListNode reverse(ListNode node){
       ListNode prev = null, now = node;
       while(now!=null){
           ListNode next = now.next;
           now.next = pre;
           pre = now;
           now = next;
       }
       return prev;
   }
}

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