LeetCode [链表] 876. Middle of the Linked List （C++和Python实现）

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 = None

class Solution:
    def middleNode(self, head: ListNode) -> ListNode:
        if not head:
            return head
        cnt = 0
        tmpHead = head
        while tmpHead:
            tmpHead = tmpHead.next
            cnt += 1
        # 这里要//整除，不然会变成浮点数，多循环一次
        cnt = cnt//2
        print(cnt)
        while cnt>0:
            cnt -= 1
            head = head.next
        return head