Reorder List

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}.

Method:

Reverse the second part of the linkedlist, and then combine them.

Solution:

Time: O(n)
Space: O(1)

/**
 * Definition for singly-linked list.
 * class ListNode {
 *     public int val;
 *     public ListNode next;
 *     ListNode(int x) { val = x; next = null; }
 * }
 */
public class Solution {
    //    s  f
    //       s    f
    // 1->2->3->4 => 3
    //    s  f
    //       s     f
    // 1->2->3->4->5 => 3
    private ListNode findMid(ListNode A) {
        ListNode slow = A;
        ListNode fast = A;
        while (fast != null && fast.next != null) {
            slow = slow.next;
            fast = fast.next.next;
        }
        return slow;
    }
    
    // 1->2->3->4
    // 1<-2->3->4
    private ListNode reverse(ListNode A) {
        ListNode prev = null;
        ListNode curr = A;
        while (curr != null) {
            ListNode next = curr.next;
            curr.next = prev;
            prev = curr;
            curr = next;
        }
        return prev;
    }
    
    // 1->2->3
    // 5->4->3
    
    // 1->2->3
    // 4->3
    public ListNode reorderList(ListNode A) {
        if (A == null || A.next == null) {
            return A;
        }
        ListNode mid = findMid(A);
        mid = reverse(mid);
        ListNode dummy = new ListNode(0);
        ListNode p = dummy;
        while (A.next != null && mid != null) {
            ListNode Anext = A.next;
            p.next = A;
            p = p.next;
            p.next = mid;
            p = p.next;
            A = Anext;
            mid = mid.next;
        }

        return dummy.next;
    }
}

/**
 * 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) {
        ListNode mid = findMid(head);
        ListNode a = head;
        ListNode b = reverse(mid);
        
        merge(a, b);
    }
    
    private void merge(ListNode a, ListNode b) {
        // 1 -> 2 -> 3
        // 5 -> 4
        ListNode pA = a;
        ListNode pB = b;
        while (pA != null && pB != null) {
            ListNode aNext = pA.next;
            ListNode bNext = pB.next;
            pA.next = pB;
            pB.next = aNext;
            pA = aNext;
            pB = bNext;
        }
    }
    
    private ListNode reverse(ListNode head) {
        //         1 -> 2 -> 3 -> 4
        // null <- 1 <- 2 -> 3 -> 4
        ListNode p = head;
        ListNode prev = null;
        while (p != null) {
            ListNode next = p.next;
            p.next = prev;
            prev = p;
            p = next;
        }
        return prev;
    }
    
    private ListNode findMid(ListNode head) {
        ListNode slow = head;
        ListNode fast = head;
        ListNode prev = null;
        while (fast != null && fast.next != null) {
            prev = slow;
            slow = slow.next;
            fast = fast.next.next;
            if (fast == null) {
                ListNode ret = slow;
                prev.next = null;
                return ret;
            }
            if (fast.next == null) {
                ListNode ret = slow.next;
                slow.next = null;
                return ret;
            }
        }
        return head;
    }
}