https://leetcode.com/problems/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.
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) { if(!head) return NULL; ListNode *slow = head; ListNode *fast = head; while(fast && fast -> next) { slow = slow -> next; fast = fast -> next -> next; } return slow; } };
又是一个快慢指针 虽然开始的时候写的有点乱 主要还是不是很清楚 $while$ 里面应该怎么写 要上课了!去占个位置回来继续写