Leetcode 876. Middle of the Linked List

Leetcode 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.

解题思路

利用快慢指针的原理，慢指针走一步，快指针走两步，当快指针走到末尾，慢指针刚好走到中间，需要留意的是，当节点数为偶数时，输出的第二个中间节点。

代码

/**
 * 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* middleNode = head; //走一步
        ListNode* fastNode = head;   //走两步
        while(fastNode->next!= NULL){
            middleNode = middleNode->next; // fastNode 比 middleNode 走得快，不需要判断middle->next != NULL
            if(fastNode->next->next != NULL) fastNode = fastNode->next->next;
            else break;  // fastNode走到头了
        }
        return middleNode;
    }
};