【leetcode】876. Middle of the Linked List

Given the head of a singly linked list, return the middle node of the linked list.

If there are two middle nodes, return the second middle node.

Example 1:

Input: head = [1,2,3,4,5]
Output: [3,4,5]
Explanation: The middle node of the list is node 3.

Example 2:

Input: head = [1,2,3,4,5,6]
Output: [4,5,6]
Explanation: Since the list has two middle nodes with values 3 and 4, we return the second one.

    本题的需求是返回一个链表的中间节点，属于那种一眼就能想到如何解答的easy题目。

    解法一，利用map存储每个节点以及节点的位置，然会中间位置的节点。

class Solution {
public:
    ListNode* middleNode(ListNode* head) {
        // return mid node 
        // cal the num of the node
        unordered_map<int,ListNode*> hash_map;
        ListNode* node=head;
        int index=0;
        while(node!=nullptr){
            hash_map[index]=node;
            node=node->next;
            index++;
        }
        return hash_map[(index/2)];
        
    }
};

     解法二，上述解法空间复杂度太高，其实还可以用快慢指针解决，快指针每个循环走两步，慢指针每个循环走一步，当快指针结束的时候，慢指针刚好走到了一半，即中间位置。（如下图所示 龟兔赛跑）

class Solution {
public:
    ListNode* middleNode(ListNode* head) {
        // 快慢指针
        ListNode* low=head;
        ListNode* fast=head;
        while(fast && fast->next){
            low=low->next;
            fast=fast->next->next;
        }
        return low;
    }
};