143 Reorder List – Medium¶
Problem:¶
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}.
Thoughts¶
This problem can be easily solved by using HashMap, but due to the requirement, “do this in-place”. So the plan is to complete in three steps. 1 Find the middle element of the list. 2 Do a reverse operation on the second half of the list. 3 Merge the first half and second half in a fashion that picks element from each in order.
Solutions:¶
/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode(int x) { val = x; }
* }
*/
public class Solution {
public void reorderList(ListNode head) {
if (head == null || head.next == null) {
return;
}
ListNode mid = findMid(head);
ListNode head2 = reverse(mid.next);
mid.next = null;
merge(head, head2);
}
private ListNode merge(ListNode head1, ListNode head2) {
ListNode fakeHead = new ListNode(-1);
ListNode node = fakeHead;
while (head1 != null && head2 != null) {
node.next = head1;
head1 = head1.next;
node.next.next = head2;
head2 = head2.next;
node = node.next.next;
}
if (head1 != null) {
node.next = head1;
}
if (head2 != null) {
node.next = head2;
}
return fakeHead.next;
}
private ListNode reverse(ListNode head) {
ListNode fakeHead = new ListNode(-1);
fakeHead.next = head;
ListNode node = head;
ListNode next = null;
while(node != null) {
next = node.next;
node.next = fakeHead.next;
fakeHead.next = node;
node = next;
}
head.next = null;
return fakeHead.next;
}
private ListNode findMid(ListNode head) {
ListNode slow = head, fast = head.next;
while (fast != null && fast.next != null) {
fast = fast.next.next;
slow = slow.next;
}
return slow;
}
}