如果 s 是字串acronym,回傳 true
在進行英文句子分割時要分割對,
「, 」(逗點)
of, and 都是重要的分割地方
--程式答案-----------------
2828. Check if a String Is an Acronym of Words
你可能只花一分鐘寫完程式碼,但是比較困難在於閱讀英文題目__________________\
class Solution {
public:
bool isAcronym(vector<string> &words, string s) {
string ws ="";
for( auto w: words){
ws += w[0];
}
if(ws == s)
return true;
else
return false;
}
};
Given an array of strings words and a string s, determine if s is an acronym of words.
The string s is considered an acronym of words if it can be formed by concatenating the first character of each string in words in order. For example, "ab" can be formed from ["apple", "banana"], but it can't be formed from ["bear", "aardvark"].
Return true if s is an acronym of words, and false otherwise.
Example 1:
Input: words = ["alice","bob","charlie"], s = "abc"
Output: true
Explanation: The first character in the words "alice", "bob", and "charlie" are 'a', 'b', and 'c', respectively. Hence, s = "abc" is the acronym.
Example 2:
Input: words = ["an","apple"], s = "a"
Output: false
Explanation: The first character in the words "an" and "apple" are 'a' and 'a', respectively.
The acronym formed by concatenating these characters is "aa".
Hence, s = "a" is not the acronym.
Example 3:
Input: words = ["never","gonna","give","up","on","you"], s = "ngguoy"
Output: true
Explanation: By concatenating the first character of the words in the array, we get the string "ngguoy".
Hence, s = "ngguoy" is the acronym.
Constraints:
1 <= words.length <= 100
1 <= words[i].length <= 10
1 <= s.length <= 100
words[i] and s consist of lowercase English letters.
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我應該是寫的不錯,才會有這麼高的人次觀看。
分享給大家。
我想到一個職場故事分享給大家,這是我親身經歷。
有一天,我被主管叫到他個人的辨公室裡,進行一對一的面談。
他說: 「Mr. 歐,我知道你很優秀,但是依照你現在的階段、經歷你要使別人更優秀」
其實,我當下聽不懂,但是,我有背起來。
過了幾年,我才理解,原來他講的是「團隊的重要」。
雖然,我是全國競賽第一名,但整個團隊的經營是主管所在意的事情。
所以,我開始寫文章,分享我的經驗,使讀者更優秀。
作者:歐育溙 Billour Ou
Version: 2023111501