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.

 

Note:

• The number of nodes in the given list will be between 1 and 100.

 

---

 

 1 /**
 2  * Definition for singly-linked list.
 3  * struct ListNode {
 4  *     int val;
 5  *     ListNode *next;
 6  *     ListNode(int x) : val(x), next(NULL) {}
 7  * };
 8  */
 9 class Solution {
10 public:
11     ListNode* middleNode(ListNode* head) {
12         if(head==NULL||head->next==NULL){
13             return head;
14         }
15         
16         ListNode* slow=head;
17         ListNode* fast=head;
18         while(fast->next&&fast->next->next){
19             fast=fast->next->next;
20             slow=slow->next;
21         }
22         
23         if(fast->next==NULL){
24             return slow;
25         }else{
26             return slow->next;
27         }
28     }
29 };

方法二：

 1 /**
 2  * Definition for singly-linked list.
 3  * struct ListNode {
 4  *     int val;
 5  *     ListNode *next;
 6  *     ListNode(int x) : val(x), next(NULL) {}
 7  * };
 8  */
 9 class Solution {
10 public:
11     ListNode* middleNode(ListNode* head) {
12         ListNode* slow=head;
13         ListNode* fast=head;
14         while(fast&&fast->next){
15             fast=fast->next->next;
16             slow=slow->next;
17         }
18         
19         return slow;
20     }
21 };

 

转载于:https://www.cnblogs.com/ruisha/p/10160568.html