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876. Middle of the Linked List [难度：简单]
【题目】
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
【题解C++】
输出中间值。最好想的办法就是遍历一遍求出链表长度，再遍历一遍，从中间处返回。
（题目说带头结点，可是我测试了下这个链表没有头结点的啊，奇怪）
/*** 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==NULL) return head;int cnt = 0;ListNode* tmpHead = head;while(tmpHead){tmpHead = tmpHead->next;cnt++;}cnt = cnt/2;while(cnt){cnt--;head = head->next;}return head;}
};【解题Python】
思路同C++。
# Definition for singly-linked list.
# class ListNode:
# def __init__(self, x):
# self.val = x
# self.next = Noneclass Solution:def middleNode(self, head: ListNode) -> ListNode:if not head:return headcnt = 0tmpHead = headwhile tmpHead:tmpHead = tmpHead.nextcnt += 1# 这里要//整除，不然会变成浮点数，多循环一次cnt = cnt//2print(cnt)while cnt>0:cnt -= 1head = head.nextreturn head这篇关于LeetCode [链表] 876. Middle of the Linked List （C++和Python实现）的文章就介绍到这儿，希望我们推荐的文章对编程师们有所帮助！
