[LeetCode] 876. Middle of the Linked List

Given a non-empty, singly linked list with head node head, return a middle node of linked list.

If there are two middle nodes, return the second middle node.

Example 1:

Input: [1,2,3,4,5]
Output: Node 3 from this list (Serialization: [3,4,5])
The returned node has value 3.  (The judge‘s serialization of this node is [3,4,5]).
Note that we returned a ListNode object ans, such that:
ans.val = 3, ans.next.val = 4, ans.next.next.val = 5, and ans.next.next.next = NULL.

Example 2:

Input: [1,2,3,4,5,6]
Output: Node 4 from this list (Serialization: [4,5,6])
Since the list has two middle nodes with values 3 and 4, we return the second one.

找单链表中点，一看到题就想到快慢指针

我的思路是开始fast, slow两个指针都在head处，fast每次走两步，slow每次走一步，再考虑到链表长度的奇偶可能导致fast走不了两步，加了个判断

ListNode* middleNode(ListNode* head) {
    if (head == NULL)
    {
        return head;
    }

    ListNode* fast = head;
    ListNode* slow = head;

    while (fast->next && fast->next->next)
    {
        fast = fast->next->next;
        slow = slow->next;
    }

    if (fast->next)
    {
        slow = slow->next;
    }

    return slow;
}

中间遇到下面这个问题，加上了几个NULL的判断就通过编译了

member access within null pointer of type ‘struct ListNode‘

再看看leetcode上其他人写的快慢指针，同样的快慢指针，这个明显要更好，不用考虑链表长度

ListNode* middleNode(ListNode* head) {
    if(head==NULL || head->next==NULL)
        return head;
    ListNode* slow=head;
    ListNode* fast=head;
    fast=fast->next->next;
    while(fast!=NULL && fast->next!=NULL){
        slow=slow->next;
        fast=fast->next->next;
    }
    return slow->next;
}

References

1. LeetCodeBug-member access within null pointer of type ‘struct ListNode‘

原文地址：https://www.cnblogs.com/arcsinw/p/9397682.html