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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