143. Reorder List

Description of Problem

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]

Constraints:

• The number of nodes in the list is in the range [1, 5 * 10^4].
• 1 <= Node.val <= 1000

Solution

Tags: LinkedList

Explanation

Simply speaking, find the middle node using fast and slow pointers, cut the list into two lists, reverse the latter and combine them.

Code (C++)

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode() : val(0), next(nullptr) {}
 *     ListNode(int x) : val(x), next(nullptr) {}
 *     ListNode(int x, ListNode *next) : val(x), next(next) {}
 * };
 */
class Solution {
public:
    void reorderList(ListNode* head) {
        ListNode* fast = head;
        ListNode* slow = head;

        // 1. Find the middle node
        while(fast->next != nullptr && fast->next->next != nullptr){
            slow = slow->next;
            fast = fast->next->next;
        }

        // 2. Cut and Reverse
        ListNode* curr = slow->next; 
        ListNode* prev = nullptr;
        slow->next = nullptr; // Cut

        // Reverse
        while(curr != nullptr){
            ListNode* temp = curr;
            curr = temp->next;
            temp->next = prev;
            prev = temp;
        }

        // 3. Merge
        ListNode* list1 = head; ListNode* list2 = prev;
        while(list1 != nullptr && list2 != nullptr){
            ListNode* temp1 = list1->next;
            ListNode* temp2 = list2->next;
            list1->next = list2;
            list2->next = temp1;
            list1 = temp1;
            list2 = temp2;
        }


    }
};

Complexity

Time Complexity:

• \(T(n)=O(n)\)
  • There are only nearly constant number of n iterations

Auxiliary Space:

• \(S(n)=O(1)\)
  • Use only constant number of variables