143. Reorder List

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]

Q5: N1 -> N2 ->N3 -> N4 -> N5 -> N6 -> ... ->Nn -> null

(convert to) (N1 -> Nn) -> (N2->N(n-1)) -> ...

N1 -> N2 -> N3 N4->N5->N6

cur1 mid

N6 N5 N4

cur2

Solution:

Step1: find middle node

Step2: reverse the 2nd half

Step3: dummyhead to append n1 n2 repeatedly

/**
 * 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) {
        // base case 
        if (head == null || head.next == null){
            return;
        }


        // 1. find the middle node
        ListNode mid = middleNode(head);
        ListNode one = head;
        ListNode two = mid.next;

        // de-link the second half from the list.

        mid.next = null;   // 这个要注意!!! 不然后续就会进入死循坏 在merge当中
      // 要记得断开

        // 2. reverse the seconde half
        // 3. merge the two halves

        merge(one, reverse(two));
    }

    private ListNode middleNode(ListNode head){
        ListNode slow = head;
        ListNode fast = head;
        while(fast.next != null && fast.next.next != null){
            slow = slow.next;
            fast = fast.next.next;
        }

        return slow;
    }

    private ListNode reverse(ListNode head){
        if (head == null || head.next == null){
            return head;
        }

        ListNode prev = null;

        while(head != null){
            ListNode next = head.next;
            head.next = prev;
            prev = head;
            head = next;
        }
        return prev;
    }

    private ListNode merge(ListNode one, ListNode two){
        ListNode dummy = new ListNode(-1);
        ListNode cur = dummy;

        while (one != null && two != null){
            cur.next = one;
            one = one.next;

            cur.next.next = two;
            two = two.next;
            cur = cur.next.next;
        }

        if (one != null){
            cur.next = one;
        }else{
            cur.next = two;
        }
        return dummy.next;
    }
}

TC: O(n)

SC: O(1)

写一题练三题!

写这题的思路 要清晰 一步一步来