Description

A sentence S is given, composed of words separated by spaces. Each word consists of lowercase and uppercase letters only.

We would like to convert the sentence to "Goat Latin" (a made-up language similar to Pig Latin.)

The rules of Goat Latin are as follows:

  • If a word begins with a vowel (a, e, i, o, or u), append "ma" to the end of the word.

    For example, the word 'apple' becomes 'applema'.

  • If a word begins with a consonant (i.e. not a vowel), remove the first letter and append it to the end, then add "ma".

    For example, the word "goat" becomes "oatgma".

  • Add one letter 'a' to the end of each word per its word index in the sentence, starting with 1.

    For example, the first word gets "a" added to the end, the second word gets "aa" added to the end and so on.

Return the final sentence representing the conversion from S to Goat Latin.

Example 1:

Input: "I speak Goat Latin"
Output: "Imaa peaksmaaa oatGmaaaa atinLmaaaaa"

Example 2:

Input: "The quick brown fox jumped over the lazy dog"
Output: "heTmaa uickqmaaa rownbmaaaa oxfmaaaaa umpedjmaaaaaa overmaaaaaaa hetmaaaaaaaa azylmaaaaaaaaa ogdmaaaaaaaaaa"

Notes:

  • S contains only uppercase, lowercase and spaces. Exactly one space between each word.
  • 1 <= S.length <= 150.

Analyse

  • 如果单词以元音(a,e,i,o,u)开头,在这个单词末尾增加"ma"

    "apple" -> "applema"

  • 如果单词以辅音开头,将单词第一个字母移动到末尾,然后在单词结尾增加"ma"

    "goat" -> "oatg" -> "oatgma"

  • 对每个单词,增加单词的序号个"a"到单词的末尾,序号从1开始

    "a b c" -> "ama bma cma" -> "amaa bmaaa cmaaaa"

简单题,直接上代码

string toGoatLatin(string S)
{
stringstream ss(S);
string tmp;
string result;
int index = 1; while (ss >> tmp)
{
if (tmp[0] == 'a' || tmp[0] == 'A' ||
tmp[0] == 'e' || tmp[0] == 'E' ||
tmp[0] == 'i' || tmp[0] == 'I' ||
tmp[0] == 'o' || tmp[0] == 'O' ||
tmp[0] == 'u' || tmp[0] == 'U')
{
tmp.append("ma");
}
else
{
string first = tmp.substr(0, 1);
tmp.erase(0, 1);
tmp.append(first + "ma");
} if (index != 1) result += " ";
result += (tmp + string(index, 'a'));
index++;
} return result;
}

[LeetCode] 824. Goat Latin的更多相关文章

  1. LeetCode 824 Goat Latin 解题报告

    题目要求 A sentence S is given, composed of words separated by spaces. Each word consists of lowercase a ...

  2. LeetCode 824. Goat Latin (山羊拉丁文)

    题目标签:String 首先把vowel letters 保存入 HashSet. 然后把S 拆分成 各个 word,遍历每一个 word: 当 word 第一个 字母不是 vowel 的时候,把第一 ...

  3. 824. Goat Latin - LeetCode

    Questioin 824. Goat Latin Solution 题目大意:根据要求翻译句子 思路:转换成单词数组,遍历数组,根据要求转换单词 Java实现: 用Java8的流实现,效率太低 pu ...

  4. 【Leetcode_easy】824. Goat Latin

    problem 824. Goat Latin solution class Solution { public: string toGoatLatin(string S) { unordered_s ...

  5. 【LeetCode】824. Goat Latin 解题报告(Python)

    作者: 负雪明烛 id: fuxuemingzhu 个人博客: http://fuxuemingzhu.cn/ 目录 题目描述 题目大意 解题方法 日期 题目地址:https://leetcode.c ...

  6. [LeetCode&Python] Problem 824. Goat Latin

    A sentence S is given, composed of words separated by spaces. Each word consists of lowercase and up ...

  7. [LeetCode] 824. Goat Latin_Easy

    A sentence S is given, composed of words separated by spaces. Each word consists of lowercase and up ...

  8. 824. Goat Latin山羊拉丁文

    [抄题]: A sentence S is given, composed of words separated by spaces. Each word consists of lowercase ...

  9. 824. Goat Latin

    class Solution { public: string toGoatLatin(string S) { S.push_back(' '); //add a space that the loo ...

随机推荐

  1. os.linesep提取当前平台使用的换行符

    1. unix平台的换行符:\n 2.DOS/Win32平台的换行符:\r\n 3.通过os.linesep函数可以提取当前所处平台的换行符,从而实现不需要关注程序运行在什么平台,也不需要根据不同的平 ...

  2. E-Explorer_2019牛客暑期多校训练营(第八场)

    题意 n个点,m条边,u,v,l,r表示点u到点v有一条边,且只有编号为\([l,r]\)的人能通过,问从点1到点n有哪些编号的人能通过 题解 先对\(l,r\)离散化,用第七场找中位数那题同样的形式 ...

  3. 牛客练习赛39 D 动态连通块+并查集 X bitset 优化

    https://ac.nowcoder.com/acm/contest/368/D 题意 小T有n个点,每个点可能是黑色的,可能是白色的.小T对这张图的定义了白连通块和黑连通块:白连通块:图中一个点集 ...

  4. 2019 Multi-University Training Contest 7

    2019 Multi-University Training Contest 7 A. A + B = C 题意 给出 \(a,b,c\) 解方程 \(a10^x+b10^y=c10^z\). tri ...

  5. bzoj 4025 二分图 lct

    题目传送门 题解: 首先关于二分图的性质, 就是没有奇环边. 题目其实就是让你判断每个时段之内有没有奇环. 其次 lct 只能维护树,(反正对于我这种菜鸟选手只会维护树), 那么对于一棵树来说, 填上 ...

  6. CodeForces 620D Professor GukiZ and Two Arrays 双指针

    Professor GukiZ and Two Arrays 题解: 将a数组都sort一遍之后, b数组也sort一遍之后. 可以观察得到 对于每一个ai来说, 整个数组bi是一个V型的. 并且对于 ...

  7. lightoj 1201 - A Perfect Murder(树形dp)

    题目链接:http://www.lightoj.com/volume_showproblem.php?problem=1201 题解:简单的树形dp,dp[0][i]表示以i为根结点不傻i的最多有多少 ...

  8. hdu 5902 GCD is Funny

    Problem Description Alex has invented a new game for fun. There are n integers at a board and he per ...

  9. 【DataBase】事务

    一.事务概述 二.事务的四大特性(ACID) 三.事务的隔离性导致的问题 四.数据库的四个隔离级别 五.数据库中的锁机制: 六.更新丢失 七.并发事务所带来的的问题 一.事务概述 事务的概念:事务是指 ...

  10. SQLi_LABS less5: GET-Double Injection - Single Quotes - String

    目录 前言 几种可能的注入方式 补充的相关知识点 前言 最近开始用SQLi_LABS学习注入,刚开始有点摸不到头脑,索性把看到的知识点记录下来,很多细节是看别人博客学的,就直接给链接了,在此向这些作者 ...