876. Middle of the 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.

 

Note:

• The number of nodes in the given list will be between 1 and 100.

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处理linked list相关题目的一个技巧是使用两个指针。两个指针（fast，slow）有两种用法：

1. fast指针每次移动两个node，slow指针每次只移动一个node。
2. fast指针先走n步

这道题可以用两个指针的第一种用法。需要注意细节，

/**
 * 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) {
        ListNode* fast = head;
        ListNode* slow = head;
        while(fast->next && fast->next->next){
            fast = fast->next->next;
            slow = slow->next;
        }
        
        if(fast->next==NULL){
            return slow;
        }
        
        return slow->next;
    }
};

 

不同的终止条件

/**
 * 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) {
        ListNode* fast = head;
        ListNode* slow = head;
        while(fast && fast->next){
            fast = fast->next->next;
            slow = slow->next;
        }

        return slow;
    }
};