876. Middle of the Linked List

经典链表题
找链表的中间节点 快慢指针

#include<iostream>
#include<vector>
#include<algorithm>
using namespace std;
//Definition for singly-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.

 struct ListNode{
     int val;
     ListNode* next;
     ListNode(int x):val(x),next(NULL){} //初始化列表
 };
 class Solution{
 public:
     ListNode* middleNode(ListNode* head){
         if(head==NULL)
             return head;
         ListNode* slow=head;
         ListNode* fast=head;
         while(fast&&fast->next){
             fast=fast->next->next;
             slow=slow->next;
         }
         return slow;
     }
 };

int main(){
    vector<int> vals={1,2,3,4,5};
    ListNode* head=NULL;
    ListNode* cur=NULL;
    for(int i=0;i<vals.size();++i){
        if(cur==NULL){
            head=cur=new ListNode(vals[i]);
        }else{
            cur->next=new ListNode(vals[i]);
            cur=cur->next;
        }
    }

    Solution solution;
    ListNode* middle=solution.middleNode(head);
    std::cout<<middle->val<<endl;
    return 0;
}