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]

Constraints:

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

From: LeetCode
Link: 143. Reorder List

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Solution:

Ideas:

1. Find the middle of the linked list: We can use the fast and slow pointer technique to find the middle node.
2. Reverse the second half of the linked list: Once we find the middle, we need to reverse the second half of the list.
3. Merge the two halves: Finally, we merge the two halves by alternating nodes from the first half and the reversed second half.

Code:

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     struct ListNode *next;
 * };
 */
void reorderList(struct ListNode* head) {
    if (!head || !head->next) return;

    // Step 1: Find the middle of the linked list
    struct ListNode *slow = head, *fast = head;
    while (fast->next && fast->next->next) {
        slow = slow->next;
        fast = fast->next->next;
    }

    // Step 2: Reverse the second half of the list
    struct ListNode *prev = NULL, *curr = slow->next, *next = NULL;
    while (curr) {
        next = curr->next;
        curr->next = prev;
        prev = curr;
        curr = next;
    }
    slow->next = NULL; // Cut the list into two halves

    // Step 3: Merge the two halves
    struct ListNode *first = head, *second = prev;
    while (second) {
        struct ListNode *tmp1 = first->next, *tmp2 = second->next;
        first->next = second;
        second->next = tmp1;
        first = tmp1;
        second = tmp2;
    }
}

// Helper function to create a new ListNode
struct ListNode* newNode(int val) {
    struct ListNode* node = (struct ListNode*)malloc(sizeof(struct ListNode));
    node->val = val;
    node->next = NULL;
    return node;
}

// Helper function to print the linked list
void printList(struct ListNode* head) {
    while (head) {
        printf("%d -> ", head->val);
        head = head->next;
    }
    printf("NULL\n");
}