[YeomChaeeun] Week 12

non-overlapping-intervals/YeomChaeeun.ts

@@ -0,0 +1,19 @@
+
+/**
+ * 알고리즘 복잡도
+ * - 시간 복잡도: O(nlogn)
+ * - 공간 복잡도: O(1)
+ */
+function eraseOverlapIntervals(intervals: number[][]): number {
+    intervals.sort((a, b) => a[1] - b[1]) 
+    let end = intervals[0][1]
+    let count = 0
+    for(let i = 1; i < intervals.length; i++) {
+        if(intervals[i][0] < end) {
+            count++
+        } else {
+            end = intervals[i][1]
+        }
+    }
+    return count
+}

remove-nth-node-from-end-of-list/YeomChaeeun.ts

@@ -0,0 +1,38 @@
+/**
+ * Definition for singly-linked list.
+ * class ListNode {
+ *     val: number
+ *     next: ListNode | null
+ *     constructor(val?: number, next?: ListNode | null) {
+ *         this.val = (val===undefined ? 0 : val)
+ *         this.next = (next===undefined ? null : next)
+ *     }
+ * }
+ */
+/**
+ * n번째 노드 제거하기
+ * 알고리즘 복잡도
+ * - 시간 복잡도: O(n)
+ * - 공간 복잡도: O(n)
+ * @param head
+ * @param n
+ */
+function removeNthFromEnd(head: ListNode | null, n: number): ListNode | null {
+  let stack: ListNode[] = [];
+  let node = head;
+
+  while (node) {
+    stack.push(node);
+    node = node.next;
+  }
+
+  // 첫 번째 노드를 제거하는 경우 추가
+  if (stack.length - n - 1 < 0) {
+    return head?.next || null;
+  }
+
+  const prevNode = stack[stack.length - n - 1];
+  prevNode.next = prevNode.next?.next || null;
+
+  return head;
+}

same-tree/YeomChaeeun.ts

@@ -0,0 +1,29 @@
+/**
+ * Definition for a binary tree node.
+ * class TreeNode {
+ *     val: number
+ *     left: TreeNode | null
+ *     right: TreeNode | null
+ *     constructor(val?: number, left?: TreeNode | null, right?: TreeNode | null) {
+ *         this.val = (val===undefined ? 0 : val)
+ *         this.left = (left===undefined ? null : left)
+ *         this.right = (right===undefined ? null : right)
+ *     }
+ * }
+ */
+/**
+ * 같은 트리인지 확인하기
+ * 알고리즘 복잡도
+ * - 시간 복잡도: O(n)
+ * - 공간 복잡도: O(n)
+ * @param p
+ * @param q
+ */
+function isSameTree(p: TreeNode | null, q: TreeNode | null): boolean {
+  if(!p || !q) {
+    return p === q;
+  }
+
+  return p.val === q.val && isSameTree(p.left, q.left) && isSameTree(p.right, q.right)
+
+}