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  • 876. Middle of the Linked List

    题目描述:


    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:

    寻找链表的中间节点,首先看看中间节点有什么特点,中间节点的到头节点的节点数等于尾节点到中间节点的节点数,或者比尾节点到中间节点的节点数多一。

    此时可以申明两个节点指向头结点,一个快节点每步走两个节点,一个慢节点每步走一个节点,当快捷点为NULL或快捷点的next节点为NULL时,慢节点指向中间节点。

    代码:

     1 /**
     2  * Definition for singly-linked list.
     3  * struct ListNode {
     4  *     int val;
     5  *     ListNode *next;
     6  *     ListNode(int x) : val(x), next(NULL) {}
     7  * };
     8  */
     9 class Solution {
    10 public:
    11     ListNode* middleNode(ListNode* head) {
    12         if (head == NULL)
    13             return head;
    14         ListNode* fast = head;
    15         ListNode* slow = head;
    16         while (fast != NULL && fast->next != NULL) {
    17             fast = fast->next;
    18             fast = fast->next;
    19             slow = slow->next;
    20         }
    21         return slow;
    22     }
    23 };
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  • 原文地址:https://www.cnblogs.com/gsz-/p/9439521.html
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