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
and100
.
思路:本题要求找单链表的中点。考虑使用快慢指针,慢指针走一步,快指针走两步,快指针走到末尾时,慢指针即指向中间节点。
C++ 代码
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode(int x) : val(x), next(NULL) {}
* };
*/
class Solution {
public:
ListNode* middleNode(ListNode* head) {
ListNode* slower = head;
ListNode* faster = head;
while (slower && faster && faster->next)
{
slower = slower->next;
faster = faster->next->next;
}
return slower;
}
};
运行时间:4ms
运行内存:8.3M
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