143. Reorder List - Medium

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.

 

first use two pointers (slow, fast) to find the middle node of the linked list;

then reverse the second half of the linked list;

then merge the two halves one by one.

p.s. 注意合并两个linked list时候的处理

time: O(n), space: O(1)

/**
 * 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;
        }
        // find middle node
        ListNode fast = head, slow = head;
        while(fast.next != null && fast.next.next != null) {
            fast = fast.next.next;
            slow = slow.next;
        }
        // split into two halves, reverse the second half
        ListNode second = reverse(slow.next);
        slow.next = null;
        // merge
        merge(head, second);
    }
    
    private ListNode reverse(ListNode head) {
        if(head == null || head.next == null) {
            return head;
        }
        ListNode prev = null, cur = head;
        while(cur != null) {
            ListNode next = cur.next;
            cur.next = prev;
            prev = cur;
            cur = next;
        }
        return prev;
    }
    
    private void merge(ListNode p1, ListNode p2) {
        while(p1 != null && p2 != null) {
            ListNode n1 = p1.next;
            ListNode n2 = p2.next;
            p1.next = p2;
            p2.next = n1;
            p1 = n1;
            p2 = n2;
        }
    }
}