876. 链表的中间结点

https://leetcode-cn.com/problems/middle-of-the-linked-list/

Java

/*
 * @Author: Goog Tech
 * @Date: 2020-07-18 22:10:42
 * @Description: https://leetcode-cn.com/problems/middle-of-the-linked-list/
 * @FilePath: \leetcode-googtech\#876. Middle of the Linked List\Solutoin.java
 */ 

 /**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */

class Solution {
    // 快指针fastNode每次走2步,慢指针slowNode每次走1步,
    // 当fastNode走到末尾时slowNode正好走到中间
    public ListNode middleNode(ListNode head) {
       ListNode fastNode = head,slowNode = head;
       while(fastNode!=null && fastNode.next!=null){
           fastNode = fastNode.next.next;
           slowNode = slowNode.next;
       } 
       return slowNode;
    }
}

Python

'''
@Author: Goog Tech
@Date: 2020-07-18 22:10:48
@Description: https://leetcode-cn.com/problems/middle-of-the-linked-list/
@FilePath: \leetcode-googtech\#876. Middle of the Linked List\Solution.py
'''

# Definition for singly-linked list.
# class ListNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.next = None

class Solution(object):
    def middleNode(self, head):
        """
        :type head: ListNode
        :rtype: ListNode
        """
        # 快指针fastNode每次走2步,慢指针slowNode每次走1步,
        # 当fastNode走到末尾时slowNode正好走到中间
        fastNode = slowNode = head
        while fastNode and fastNode.next:
            fastNode = fastNode.next.next
            slowNode = slowNode.next
        return slowNode