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 liststruct 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 liststruct 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 halvesstruct 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");
}