Leetcode

876. Middle of the Linked List

by hooni posted May 04, 2020
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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`.

```/**
* public class ListNode {
*     int val;
*     ListNode next;
*     ListNode() {}
*     ListNode(int val) { this.val = val; }
*     ListNode(int val, ListNode next) { this.val = val; this.next = next; }
* }
*/
class Solution {
int count = 0;

while(node != null){
count++;
node = node.next;
}

for(int i=0; i<count/2; i++){
node = node.next;
}

return node;
}
}```

```/**
* public class ListNode {
*     int val;
*     ListNode next;
*     ListNode() {}
*     ListNode(int val) { this.val = val; }
*     ListNode(int val, ListNode next) { this.val = val; this.next = next; }
* }
*/
class Solution {
List<ListNode> list = new ArrayList<>();
int count = 0;

while(node != null){
count++;
node = node.next;
}

return list.get(count/2);
}
}```