LeetCode #876 MiddleOfTheLinkedList. Algorithm,알고리즘,LeetCode,Codefights,CodeSignal,코드파이트,코드시그널,예제,그래프,Graph,example,c++,java,재귀,recursive,datastructure,techinterview,coding,코딩인터뷰,기술면접
Runtime: 0 ms, faster than 100.00% of C++ online submissions for Middle of the Linked List.
Memory Usage: 8.5 MB, less than 67.80% of C++ online submissions for Middle of the Linked List.
LeetCode #876
Q.
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.
주어진 비어있지 않은, 싱글 연결리스트의 헤드 노드가 있고, 그 연결리스트의 중간 노드를 반환해라.
두개의 중간 노드가 있으면, 두번째 중간 노드를 반환해라.
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.
Process
// Process
//1. Input head node
//2. Count length of linkedList
//3. Find middle of linkedList
//4. Return middle node of linkedList
// 처리과정
//1. 헤드 노드를 입력받는다.
//2. 연결리스트의 길이를 센다.
//3. 연결리스트의 중간을 찾는다.
//4. 중간 노드를 반환한다.
Code.. lemme see example code!!!
코드.. 예제코드를 보자!!!
/**
* 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) {
int length = 1;
// 2.
ListNode *temp = head;
while (temp->next != 0) {
++length;
temp = temp->next;
}
// 3.
for (int i = 0; i < length / 2; ++i) {
head = head->next;
}
// 4.
return head;
}
};
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