Given a singly linked list L: L0→L1→…→Ln-1→Ln,
reorder it to: L0→Ln→L1→Ln-1→L2→Ln-2→…
You may not modify the values in the list's nodes, only nodes itself may be changed.
Example 1:
Given 1->2->3->4, reorder it to 1->4->2->3.
Example 2:
Given 1->2->3->4->5, reorder it to 1->5->2->4->3.
代码
Approach #1 fast-slow pointer
public class Solution {
public void reorderList(ListNode head) {
if (head == null || head.next == null) return;
// find middle
ListNode slow = head, fast = head; // fast = head.next 也能通过！
while (fast != null && fast.next != null){
slow = slow.next;
fast = fast.next.next;
}
ListNode rHead = slow.next;
slow.next = null;
// reverse ListNode on the right side
ListNode prev = null;
while (rHead != null) {
ListNode temp = rHead.next;
rHead.next = prev;
prev = rHead;
rHead = temp;
}
// merge two list
rHead = prev;
ListNode lHead = head;
while (lHead != null && rHead != null){
ListNode next = lHead.next;
// add right head
lHead.next = rHead;
rHead = rHead.next;
// update left head
lHead.next.next = next;
lHead = next;
}
}
}