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.
题目分析:
本题要求中间节点，那么最直接的思路就是先从头遍历该链表，统计出链表长度n，那么中间的节点就应该是第(n+1)/2个，那么在从列表头往后遍历到第(n+1)/2个就是所求的中间节点。
虽然这种方法的时间复杂度为O(n)，但是我们需要两次遍历该链表。
一种比较方便的做法是利用快慢指针，这也是在链表的问题中比较常见的一种策略。在本题中慢指针每次前进一步，快指针每次前进两步，这样在快指针到达链表末尾的时候，慢指针就会指向中间节点。这里注意快慢指针在一开始时都应该初始化为头结点。这种做法相比于第一种直观做法，虽然时间复杂度还是O(n)，但算法只需要一次列表遍历就可以了。
对应的Golang代码如下
/**
* Definition for singly-linked list.
* type ListNode struct {
* Val int
* Next *ListNode
* }
*/
func middleNode(head *ListNode) *ListNode {
if head == nil || head.Next == nil {
return head
}
var slow *ListNode = head
var fast *ListNode = head
for fast != nil && fast.Next != nil {
fast = fast.Next.Next
slow = slow.Next
}
return slow
}
对应的C#代码如下
/**
* Definition for singly-linked list.
* public class ListNode {
* public int val;
* public ListNode next;
* public ListNode(int x) { val = x; }
* }
*/
public class Solution {
public ListNode MiddleNode(ListNode head) {
if (head == null || head.next == null) return head;
var slow = head;
var fast = head;
while (fast != null && fast.next != null)
{
slow = slow.next;
fast = fast.next.next;
}
return slow;
}
}
对应的C++代码如下
/**
* 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) {
if (head == NULL || head->next == NULL) return head;
ListNode *slow = head, *fast = head;
while (fast != NULL && fast->next != NULL) {
slow = slow->next;
fast = fast->next;
if (fast != NULL) fast = fast->next;
}
return slow;
}
};
对应的java代码如下
/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode(int x) { val = x; }
* }
*/
class Solution {
public ListNode middleNode(ListNode head) {
ListNode slow = head, fast = head;
while (fast != null && fast.next != null) {
slow = slow.next;
fast = fast.next;
if (fast.next != null) fast = fast.next;
}
return slow;
}
}
