はじめに
以下のサイトの問題の解法と関連知識をまとめる。
問題
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.
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
word1andword2consist of lowercase English letters.
回答
class Solution { public String mergeAlternately(String word1, String word2) { StringBuilder result = new StringBuilder(word1.length() + word2.length()); int i = 0; while(i < word1.length() || i < word2.length()) { if (i < word1.length()) { result.append(word1.charAt(i)); } if (i < word2.length()) { result.append(word2.charAt(i)); } i++; } return result.toString(); } }
アプローチ
- 文字列を格納するStringBuilderインスタンスを生成する。StringBuilderインスタンスの初期容量も合わせて指定することで文字列の追加のたびに容量拡張を行わなくて済むためパフォーマンス効率が良くなると考えた。スレッドセーフでなくても問題ないと考えたので、パフォーマンスが良いStringBuilderを採用。
- 文字列の長さ分ループを回し、ループの中で文字列1と文字列2を交互に追加する。文字列1もしくは文字列2が存在し続ける限りループを回す。
結果
Appendix
- StringBuilder vs StringBuffer
- ライブラリの仕様
