Add: Add 2025/2/3

Author: whats2000

.github/workflows/static.yml

@@ -30,7 +30,7 @@ jobs:
     runs-on: ubuntu-latest
     steps:
       - name: Checkout
-        uses: actions/checkout@v3
+        uses: actions/checkout@v4
       - name: Setup Pages
         uses: actions/configure-pages@v3
       - name: Upload artifact

3105-Longest Strictly Increasing or Strictly Decreasing Subarray/Note.md

@@ -0,0 +1,66 @@
+# 3105. Longest Strictly Increasing or Strictly Decreasing Subarray
+
+You are given an array of integers `nums`. Return the length of the longest subarray of `nums` 
+which is either strictly increasing or strictly decreasing.
+
+## 基礎思路
+我們可以觀察兩個元素之間差異，來決定趨勢，趨勢共有三種可能:
+- 變換趨勢: 重置長度為 2 (因為檢測到趨勢變換時已經有兩個元素)
+- 維持趨勢: 此時將當前趨勢長度加一即可
+- 保持不變: 當數字相同時，我們需要紀錄當前長度為 1 (因為題目要求嚴格遞增或遞減)
+
+## 解題步驟
+
+### Step 1: 初始化變數
+
+```typescript
+const n = nums.length;
+
+let maxLength = 1;
+let currentLength = 1;
+// 當前 subarray 的趨勢
+// 1 代表遞增
+// -1 代表遞減
+// 0 代表無趨勢 (或重置)
+let currentType = 0;
+```
+
+### Step 2: 迭代數組
+
+```typescript
+for (let i = 1; i < n; i++) {
+  const different = nums[i] - nums[i - 1];
+  // 檢查趨勢類型
+  const newType = different > 0 ? 1 : different < 0 ? -1 : 0;
+
+  if (newType === 0) {
+    // 當前元素與前一個元素相同，重置 subarray 長度為 1
+    currentLength = 1;
+    currentType = 0;
+  } else if (newType === currentType) {
+    // 當前元素與前一個元素趨勢相同，增加 subarray 長度
+    currentLength++;
+  } else {
+    // 當前元素與前一個元素趨勢不同，重置 subarray 長度為 2，開始新的 subarray 計算
+    currentLength = 2;
+    currentType = newType;
+  }
+
+  // 更新最大長度
+  if (currentLength > maxLength) {
+    maxLength = currentLength;
+  }
+}
+```
+
+## 時間複雜度
+- 計算趨勢的時間複雜度為 $O(n)$
+- 總體時間複雜度為 $O(n)$
+
+> $O(n)$
+
+## 空間複雜度
+- 僅使用常數空間
+- 總空間複雜度為 $O(1)$
+
+> $O(1)$

3105-Longest Strictly Increasing or Strictly Decreasing Subarray/answer.ts

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+function longestMonotonicSubarray(nums: number[]): number {
+  const n = nums.length;
+
+  let maxLength = 1;
+  let currentLength = 1;
+  // The type of the current subarray:
+  // 1 for increasing,
+  // -1 for decreasing,
+  // 0 for none (or reset)
+  let currentType = 0;
+
+  for (let i = 1; i < n; i++) {
+    const different = nums[i] - nums[i - 1];
+    // Determine the new trend
+    const newType = different > 0 ? 1 : different < 0 ? -1 : 0;
+
+    if (newType === 0) {
+      // Reset when equal.
+      currentLength = 1;
+      currentType = 0;
+    } else if (newType === currentType) {
+      // Continue in the same direction.
+      currentLength++;
+    } else {
+      // New trend: start a new subarray that includes the previous element.
+      currentLength = 2;
+      currentType = newType;
+    }
+
+    // Update the maximum length.
+    if (currentLength > maxLength) {
+      maxLength = currentLength;
+    }
+  }
+
+  return maxLength;
+}