876. Middle of the Linked List

tags: Easy、Linked List

Given the head of a singly linked list, return the middle node of the linked list.
If there are two middle nodes, return the second middle node.

解法:

將原本的fast->next->next;改成使用fast->next;寫兩次，有些編譯器可以更好的去改善效能跟速度，所以這樣寫的速度會更快

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     struct ListNode *next;
 * };
 */
struct ListNode* middleNode(struct ListNode* head) {
    struct ListNode *slow = head;
    struct ListNode *fast = head;
    
    while (fast != NULL && fast->next != NULL) {
        slow = slow->next;
        fast = fast->next;
        fast = fast->next;
    }

    return slow;
}