add post '(Leetcode) 19 - Remove Nth Node From End of List'

Author: dev-jonghoonpark

_posts/2024-06-10-leetcode-19.md

@@ -0,0 +1,66 @@
+---
+layout: post
+title: (Leetcode) 19 - Remove Nth Node From End of List
+categories: [스터디-알고리즘]
+tags:
+  [
+    자바,
+    java,
+    리트코드,
+    Leetcode,
+    알고리즘,
+    algorithm,
+    linked list,
+    index,
+    node,
+  ]
+date: 2024-06-10 01:00:00 +0900
+toc: true
+---
+
+기회가 되어 [달레님의 스터디](https://github.com/DaleStudy/leetcode-study)에 참여하여 시간이 될 때마다 한문제씩 풀어보고 있다.
+
+[https://neetcode.io/practice](https://neetcode.io/practice)
+
+---
+
+[https://leetcode.com/problems/remove-nth-node-from-end-of-list/](https://leetcode.com/problems/remove-nth-node-from-end-of-list/)
+
+## 내가 작성한 풀이
+
+처음에는 linked list 의 길이를 재고, 어떤 node를 삭제해야하는지 확인하여 삭제하도록 하였다.
+
+```java
+class Solution {
+    public ListNode removeNthFromEnd(ListNode head, int n) {
+		ListNode current = head;
+        int length = 0;
+        while(current != null) {
+            length++;
+            current = current.next;
+        }
+
+        if (length == 1) {
+            return null;
+        }
+
+        if (length == n) {
+			return head.next;
+		}
+
+        int removeIndex = length - n;
+        int pointer = 0;
+        current = head;
+        while (pointer < removeIndex - 1) {
+            current = current.next;
+            pointer++;
+        }
+        current.next = current.next.next;
+        return head;
+	}
+}
+```
+
+### TC, SC
+
+시간 복잡도는 O(n)이고, 공간 복잡도는 O(1)이다.