0876. Middle of the Linked List

876. 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.

题目大意 #

输出链表中间结点。这题在前面题目中反复出现了很多次了。

如果链表长度是奇数,输出中间结点是中间结点。如果链表长度是双数,输出中间结点是中位数后面的那个结点。

解题思路 #

这道题有一个很简单的做法,用 2 个指针只遍历一次就可以找到中间节点。一个指针每次移动 2 步,另外一个指针每次移动 1 步,当快的指针走到终点的时候,慢的指针就是中间节点。

代码 #


package leetcode

/**
 * 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
	}
	p1 := head
	p2 := head
	for p2.Next != nil && p2.Next.Next != nil {
		p1 = p1.Next
		p2 = p2.Next.Next
	}
	length := 0
	cur := head
	for cur != nil {
		length++
		cur = cur.Next
	}
	if length%2 == 0 {
		return p1.Next
	}
	return p1
}


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