LeetCode 876. 链表的中间结点 简单

我的解题

按照 labuladong 双指针技巧汇总 中，给出的思路编写的code。

#include <bits/stdc++.h>
using namespace std;
/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode() : val(0), next(nullptr) {}
 *     ListNode(int x) : val(x), next(nullptr) {}
 *     ListNode(int x, ListNode *next) : val(x), next(next) {}
 * };
 */

struct ListNode
{
    int val;
    ListNode *next;
    ListNode() :
                    val(0), next(nullptr)
    {
    }
    ListNode(int x) :
                    val(x), next(nullptr)
    {
    }
    ListNode(int x, ListNode *next) :
                    val(x), next(next)
    {
    }
};
class Solution
{
public:
    ListNode* middleNode(ListNode *head)
    {
        ListNode *fast = head;
        ListNode *slow = head;
        while (fast && fast->next)
        {
            fast = fast->next->next;
            slow = slow->next;
        }
        return slow;
    }
};
int main()
{
    Solution solu;
}