Description
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.
Note:
- The number of nodes in the given list will be between 1 and 100.
Analyse
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;
}