Merge branch 'main' of https://github.com/JuneLin2001/problem-solving

Author: JuneLin2001

docs/LeetCode/717. 1-bit and 2-bit Characters/README.mdx

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+---
+title: 717. 1-bit and 2-bit Characters
+description: We have two special characters:The first character can be represented by one bit 0. The second character can be represented by two bits (10 or 11). Given a binary array bits that ends with 0, return true if the last character must be a one-bit character.
+keywords: [LeetCode, 717. 1-bit and 2-bit Characters, Easy, Array]
+tags: [LeetCode, Easy, Array]
+---
+
+# [{frontMatter.title}](https://leetcode.com/problems/1-bit-and-2-bit-characters/)
+
+import Tabs from "@theme/Tabs";
+import TabItem from "@theme/TabItem";
+import CodeBlock from "@theme/CodeBlock";
+import DifficultyBadge from "@site/src/components/Badges/DifficultyBadge";
+import Solution from "!!raw-loader!./solution.js";
+import Description from "./_Description.md";
+import Examples from "./_Examples.md";
+
+<DifficultyBadge difficulty="Easy" />
+
+<Tabs>
+  <TabItem value="description" label="題目描述" default>
+    <details open>
+      <summary>Description</summary>
+        <Description />
+        <Examples />
+    </details> 
+  </TabItem>
+
+  <TabItem value="solution" label="解答">
+   ## Solution
+    <CodeBlock language="js">{Solution}</CodeBlock>
+  </TabItem>
+</Tabs>
+
+## 解題思路
+
+題目給一組只由 `0` 和 `1` 所組成的陣列，要判斷最後一位的 `0` 是否是單獨存在，而不是接續前面的 `[1,0]` 這樣的組合  
+最後一位一定會是 `0`，所以只需要到 `bits.length - 1` 就行  
+因為 return 時候會用到 `i` 來判斷結果，故不使用 for 選擇使用 while 迴圈來遍歷陣列
+
+```js
+var isOneBitCharacter = function (bits) {
+  let i = 0;
+  while (i < bits.length - 1) {
+    if (bits[i] === 0) {
+      // 當前是 0 的話判斷下一個
+      i++;
+    } else {
+      // 不是 0 就代表當前是 1 ，且 1 一定會組成 [1,0] 或 [1,1]，所以 += 2
+      i += 2;
+    }
+  }
+
+  return i === bits.length - 1;
+  // 如果 i === bits.length - 1 的話就代表只剩下一個 0 ，反之則否
+};
+```
+
+## 心得
+
+題目描述超不直觀，花了一點時間才看懂

docs/LeetCode/717. 1-bit and 2-bit Characters/_Description.md

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+We have two special characters:
+
+- The first character can be represented by one bit `0`.
+- The second character can be represented by two bits (`10` or `11`).
+
+Given a binary array `bits` that ends with `0`, return `true` if the last character must be a one-bit character.

docs/LeetCode/717. 1-bit and 2-bit Characters/_Examples.md

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+#### Example 1:
+
+> **Input:** bits = [1,0,0]  
+> **Output:** true  
+> **Explanation:** The only way to decode it is two-bit character and one-bit character.
+> So the last character is one-bit character.
+
+#### Example 2:
+
+> **Input:** bits = [1,1,1,0]  
+> **Output:** false  
+> **Explanation:** he only way to decode it is two-bit character and two-bit character.
+> So the last character is not one-bit character.
+
+## Constraints:
+
+- `1 <= bits.length <= 1000`
+- `bits[i]` is either `0` or `1`.

docs/LeetCode/717. 1-bit and 2-bit Characters/solution.js

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+/**
+ * @param {number[]} bits
+ * @return {boolean}
+ */
+var isOneBitCharacter = function (bits) {
+  let i = 0;
+  while (i < bits.length - 1) {
+    if (bits[i] === 0) {
+      i++;
+    } else {
+      i += 2;
+    }
+  }
+
+  return i === bits.length - 1;
+};