254th Commit

README.md

@@ -60,19 +60,20 @@ Ace Coding Interview with 75 Qs
 <details>
   <summary>Problems</summary>
 
-| Array / String                                 |                  |        |
-| ---------------------------------------------- | ---------------- | ------ |
-| 1768. Merge Strings Alternately                | [Solution][1768] | Easy   |
-| 1071. Greatest Common Divisor of Strings       | [Solution][1071] | Easy   |
-| 1431. Kids With the Greatest Number of Candies | [Solution][1431] | Easy   |
-| 605. Can Place Flowers                         | [Solution][605]  | Easy   |
-| 345. Reverse Vowels of a String                | [Solution][345]  | Easy   |
-| 151. Reverse Words in a String                 | [Solution][151]  | Medium |
-| 238. Product of Array Except Self              | [Solution][238]  | Medium |
-| 334. Increasing Triplet Subsequence            | [Solution][334]  | Medium |
-| 443. String Compression                        | [Solution][443]  | Medium |
+| Array / String                                 |                  |        |                          |
+| ---------------------------------------------- | ---------------- | ------ | ------------------------ |
+| 1768. Merge Strings Alternately                | [Solution][1768] | Easy   | [詳解][1768-explanation] |
+| 1071. Greatest Common Divisor of Strings       | [Solution][1071] | Easy   | 詳解                     |
+| 1431. Kids With the Greatest Number of Candies | [Solution][1431] | Easy   | 詳解                     |
+| 605. Can Place Flowers                         | [Solution][605]  | Easy   | 詳解                     |
+| 345. Reverse Vowels of a String                | [Solution][345]  | Easy   | 詳解                     |
+| 151. Reverse Words in a String                 | [Solution][151]  | Medium | 詳解                     |
+| 238. Product of Array Except Self              | [Solution][238]  | Medium | 詳解                     |
+| 334. Increasing Triplet Subsequence            | [Solution][334]  | Medium | 詳解                     |
+| 443. String Compression                        | [Solution][443]  | Medium | 詳解                     |
 
 [1768]: ./src/page-17/1768.%20Merge%20Strings%20Alternately/mergeAlternately.ts
+[1768-explanation]: ./src/page-17/1768.%20Merge%20Strings%20Alternately/README.md
 [1071]: ./src/page-11/1071.%20Greatest%20Common%20Divisor%20of%20Strings/gcdOfStrings.ts
 [1431]: ./src/page-14/1431.%20Kids%20With%20the%20Greatest%20Number%20of%20Candies/kidsWithCandies.ts
 [605]: ./src/page-6/605.%20Can%20Place%20Flowers/canPlaceFlowers.ts

src/page-17/1768. Merge Strings Alternately/README.md

@@ -0,0 +1,87 @@
+# 1768. 交替合併字串 (Merge Strings Alternately)
+
+給你兩個字串 `word1` 和 `word2`。請你從 `word1` 開始，透過交替加上字母來合併字串。如果一個字串比另一個字串長，就將多出來的字母追加到合併後字串的結尾。
+
+返回*合併後的字串*。
+
+範例 1：
+
+```ts
+輸入: word1 = "abc", word2 = "pqr"
+輸出: "apbqcr"
+說明: 字串合併情況如下所示：
+word1: a   b   c
+word2:   p   q   r
+合併後: a p b q c r
+```
+
+範例 2：
+
+```ts
+輸入: word1 = "ab", word2 = "pqrs"
+輸出: "apbqrs"
+說明: 注意，word2 比 word1 長，"rs" 需要追加到合併後字串的結尾。
+word1: a   b
+word2:   p   q   r   s
+合併後: a p b q   r   s
+```
+
+範例 3：
+
+```ts
+輸入: word1 = "abcd", word2 = "pq"
+輸出: "apbqcd"
+說明: 注意，word1 比 word2 長，"cd" 需要追加到合併後字串的結尾。
+word1: a   b   c   d
+word2:   p   q
+合併後: a p b q c   d
+```
+
+## 解題
+
+```ts
+export const mergeAlternately: MergeAlternately = (word1, word2) => {
+  const chars1 = Array.from(word1);
+  const chars2 = Array.from(word2);
+
+  // 使用 flatMap 方法，將兩個字串的字元交替合併
+  let mergedChars = chars1.flatMap((char, index) => [char, chars2[index]]);
+
+  // 如果 chars2 比 chars1 長，說明還有剩下的字元需要附加到結果陣列中
+  if (chars2.length > chars1.length) {
+    mergedChars = [...mergedChars, ...chars2.slice(chars1.length)];
+  }
+
+  return mergedChars.join('');
+};
+```
+
+## 解題 2
+
+使用雙指標 (Two Pointers)
+
+```ts
+export const mergeAlternately2: MergeAlternately = (word1, word2) => {
+  let result = ''; // 儲存合併後的結果
+
+  let pointer1 = 0; // 指向 word1 的當前位置
+  let pointer2 = 0; // 指向 word2 的當前位置
+
+  // 當兩個指標都還在有效範圍時，交替加入字元
+  while (pointer1 < word1.length || pointer2 < word2.length) {
+    if (pointer1 < word1.length) {
+      // 從 word1 中取一個字元並加入結果，然後遞增指標
+      result += word1[pointer1];
+      pointer1 += 1;
+    }
+
+    if (pointer2 < word2.length) {
+      // 從 word2 中取一個字元並加入結果，然後遞增指標
+      result += word2[pointer2];
+      pointer2 += 1;
+    }
+  }
+
+  return result;
+};
+```

src/page-17/1768. Merge Strings Alternately/mergeAlternately.ts

@@ -4,16 +4,16 @@ type MergeAlternately = (word1: string, word2: string) => string;
  * Accepted
  */
 export const mergeAlternately: MergeAlternately = (word1, word2) => {
-  const array1 = Array.from(word1);
-  const array2 = Array.from(word2);
+  const chars1 = Array.from(word1);
+  const chars2 = Array.from(word2);
 
-  let zip = array1.flatMap((cv, i) => [cv, array2[i]]);
+  let mergedChars = chars1.flatMap((char, index) => [char, chars2[index]]);
 
-  if (array2.length > array1.length) {
-    zip = [...zip, ...array2.slice(array1.length)];
+  if (chars2.length > chars1.length) {
+    mergedChars = [...mergedChars, ...chars2.slice(chars1.length)];
   }
 
-  return zip.join('');
+  return mergedChars.join('');
 };
 
 /**
@@ -22,18 +22,18 @@ export const mergeAlternately: MergeAlternately = (word1, word2) => {
 export const mergeAlternately2: MergeAlternately = (word1, word2) => {
   let result = '';
 
-  let w1i = 0;
-  let w2i = 0;
+  let pointer1 = 0;
+  let pointer2 = 0;
 
-  while (w1i < word1.length || w2i < word2.length) {
-    if (w1i < word1.length) {
-      result += word1[w1i];
-      w1i += 1;
+  while (pointer1 < word1.length || pointer2 < word2.length) {
+    if (pointer1 < word1.length) {
+      result += word1[pointer1];
+      pointer1 += 1;
     }
 
-    if (w2i < word2.length) {
-      result += word2[w2i];
-      w2i += 1;
+    if (pointer2 < word2.length) {
+      result += word2[pointer2];
+      pointer2 += 1;
     }
   }