LeetCode 876. Middle of the Linked List

原题链接在这里：https://leetcode.com/problems/middle-of-the-linked-list/

题目：

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.

题解：

walker每次跳一步. runner每次跳两步. runner 跳到尾时, walker就是中点.

list偶数长度时, 中点找中间两个前者时, while的终止条件到是(runner != null && runner.next != null).

找后者时, while的终止条件是(runner != null && runner.next != null).

Time Complexity: O(n).

Space: O(1).

AC Java:

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
class Solution {
    public ListNode middleNode(ListNode head) {
        if(head == null || head.next == null){
            return head;
        }
        
        ListNode walker = head;
        ListNode runner = head;
        while(runner != null && runner.next != null){
            walker = walker.next;
            runner = runner.next.next;
        }
        
        return walker;
    }
}

AC C++:

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode() : val(0), next(nullptr) {}
 *     ListNode(int x) : val(x), next(nullptr) {}
 *     ListNode(int x, ListNode *next) : val(x), next(next) {}
 * };
 */
class Solution {
public:
    ListNode* middleNode(ListNode* head) {
        if(!head || !head->next){
            return head;
        }

        ListNode* walker = head;
        ListNode* runner = head;
        while(runner && runner->next){
            walker = walker->next;
            runner = runner->next->next;
        }

        return walker;
    }
};

AC Python:

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution:
    def middleNode(self, head: Optional[ListNode]) -> Optional[ListNode]:
        walker = head
        runner = head
        while runner and runner.next:
            walker = walker.next
            runner = runner.next.next
        return walker

跟上 Delete the Middle Node of a Linked List.