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