1768.Merge Strings Alternately === ###### tags: `Easy`,`String`,`Two Pointers` [1768. Merge Strings Alternately](https://leetcode.com/problems/merge-strings-alternately/) ### 題目描述 You are given two strings `word1` and `word2`. Merge the strings by adding letters in alternating order, starting with `word1`. If a string is longer than the other, append the additional letters onto the end of the merged string. Return *the merged string*. ### 範例 **Example 1:** ``` Input: word1 = "abc", word2 = "pqr" Output: "apbqcr" Explanation: The merged string will be merged as so: word1: a b c word2: p q r merged: a p b q c r ``` **Example 2:** ``` Input: word1 = "ab", word2 = "pqrs" Output: "apbqrs" Explanation: Notice that as word2 is longer, "rs" is appended to the end. word1: a b word2: p q r s merged: a p b q r s ``` **Example 3:** ``` Input: word1 = "abcd", word2 = "pq" Output: "apbqcd" Explanation: Notice that as word1 is longer, "cd" is appended to the end. word1: a b c d word2: p q merged: a p b q c d ``` **Constraints**: * 1 <= `word1.length`, `word2.length` <= 100 * `word1` and `word2` consist of lowercase English letters. ### 解答 #### Javascript ```javascript= function mergeAlternately(word1, word2) { const result = []; for (let i = 0; i < 100; i++) { if (!word1[i] && !word2[i]) break; result.push(word1[i] ?? ''); result.push(word2[i] ?? ''); } return result.join(''); } ``` > [name=Marsgoat][time=Apr 18, 2023] #### Python ```python= class Solution: def mergeAlternately(self, word1: str, word2: str) -> str: ans = [] for ch1, ch2 in zip_longest(word1, word2, fillvalue=''): ans.append(ch1 + ch2) return "".join(ans) ``` 上面邏輯縮成一行 ```python= class Solution: def mergeAlternately(self, word1: str, word2: str) -> str: return "".join(ch1 + ch2 for ch1, ch2 in zip_longest(word1, word2, fillvalue='')) ``` > [name=Ron Chen][time=Tue, Apr 18, 2023] #### Java 練習 Java Stream API ```java= class Solution { public String mergeAlternately(String word1, String word2) { StringBuilder sb = new StringBuilder(); IntStream.range(0, Math.min(word1.length(), word2.length())) .forEach(i -> sb.append(word1.charAt(i)) .append(word2.charAt(i))); if(word1.length() > word2.length()) { sb.append(word1.substring(word2.length())); } else { sb.append(word2.substring(word1.length())); } return sb.toString(); } } ``` > [name=Ron Chen][time=Tue, Apr 18, 2023] #### TypeScript ```typescript= function mergeAlternately(word1: string, word2: string): string { let result = ''; if (word1.length >= word2.length) { for (let i = 0; i < word1.length; i++) { result += word1.charAt(i); result += word2.charAt(i); } } else { for (let i = 0; i < word2.length; i++) { result += word1.charAt(i); result += word2.charAt(i); } } return result; } ``` > Note: `String.prototype.charAt()` 如果超過索引值會回傳空字串 > [name=Sheep][time=Apr 18, 2023] ### Reference [回到題目列表](https://hackmd.io/@Marsgoat/leetcode_every_day)