C语言 统计歌词中的单词个数，并找出最短和最长的单词(必须用指针)求帮忙，感谢！！！

Posted 2023-04-06

统计歌词中的单词个数，并找出最短和最长的单词。
歌词：Daddy finger, Daddy finger, where are you? Here I am. Here I am. How do you do? Mammy finger, Mammy finger, where are you? Here I am. Here I am. How do you do? Brother finger, Brother finger, where are you? Here I am. Here I am. How do you do? Sister finger, Sister finger, where are you? Here I am. Here I am. How do you do? Baby finger, Baby finger, where are you? Here I am. Here I am. How do you do?
设计函数实现：
函数int gecifun( char *geci, char *strmax, char *strmin)——实现求geci中的单词个数并返回，strmax和strmin中分别存放最长和最短单词；
主函数——调用函数gecifun并输出。
程序输出：歌词中共有85个单词。
歌词中最长的单词为Brother，最短的单词为I。

参考技术A 函数int gecifun( char *geci, char *strmax, char *strmin)——实现求geci中的单词个数并返回，strmax和strmin中分别存放最长和最短单词；
主函数——调用函数gecifun并输出。 参考技术B #include <stdio.h>
#include <stdlib.h>
#include <string.h>
#define N 500
int gecifun(char *geci, char *strmax, char *strmin)

char temp[30];//存储每个单词数组
int num=0,i,word=0;//单词数量和每个单词数组下标
for(i=0;(geci[i])!='\0';i++)

if(geci[i]==' '||geci[i]==',' || geci[i]=='.' || geci[i]=='?') //单词分隔符


if (word != 0)
temp[word]='\0';
if (num==1)//获取第一个单词的时候，将第一个单词复制给最大和最小
strcpy(strmax,temp);
strcpy(strmin,temp);

else//第二个单词之后比较字符串长度之后进行复制


if (strlen(temp)>=strlen(strmax))
strcpy(strmax,temp);
if (strlen(temp)<=strlen(strmin))
strcpy(strmin,temp);

num++;//单词数量递增
word = 0;//单词数组下标置零


else
temp[word++] = geci[i];//每个单词按位置赋值

return num;

int main()

int acount;
char lstr[30],sstr[30];
char song[N] = "Daddy finger, Daddy finger, where are you? Here I am. Here I am."
"How do you do? Mammy finger, Mammy finger, where are you? Here I am. Here "
"I am. How do you do? Brother finger, Brother finger, where are you? Here "
"I am. Here I am. How do you do? Sister finger, Sister finger, where are you?"
"Here I am. Here I am. How do you do? Baby finger, Baby finger, where are you?"
"Here I am. Here I am. How do you do?";
acount=gecifun(song,lstr,sstr);
printf("歌词中共有%d个单词。歌词中最长的单词为%s，最短的单词为%s。\n",acount,lstr,sstr);
return 0;

最短和最长的城市名称

【中文标题】最短和最长的城市名称

【英文标题】：shortest and longest city name

【发布时间】：2019-12-25 06:49:03

【问题描述】：

查询 STATION 中两个城市的 CITY 名称最短和最长，以及它们各自的长度（即：名称中的字符数）。如果有不止一个最小或最大的城市，请选择按字母顺序排列的第一个城市。

请问这个问题的答案是什么？

  SELECT CITY,
         LENGTH(CITY)
    FROM STATION
ORDER BY CITY ASC
   FETCH FIRST 1 ROWS ONLY ;

为什么此查询不适用于获取最短长度城市？

【问题讨论】：

这不起作用，因为您是通过city 订购的，而您可能打算通过length(city) 订购。

【参考方案1】：

我会在这里使用ROW_NUMBER：

WITH cte AS (
    SELECT s.*, ROW_NUMBER() OVER (ORDER BY LENGTH(CITY), CITY) rn_first,
        ROW_NUMBER() OVER (ORDER BY LENGTH(CITY) DESC, CITY) rn_last
    FROM STATION s
)

SELECT CITY, LENGTH(CITY) AS CITY_LENGTH
FROM cte
WHERE 1 IN (rn_first, rn_last)
ORDER BY LENGTH(CITY);

这将生成一个包含两个记录的结果集，最短的城市名称出现在最前面，最长的出现在最后。

【讨论】：

是的，我在黑客等级上使用 row_number 做到了这一点。Oracle 也只接受获取前 1 行，但这在黑客等级上不起作用。

【参考方案2】：

我最喜欢在 Oracle 中执行此操作的方法是使用 keep 关键字：

select min(city) keep (dense_rank first order by length(city) asc) as shortest_city,
       min(city) keep (dense_rank first order by length(city) desc) as longest_city,
       min(length(city)) as min_len,
       max(length(city)) as max_len
from station;

根据我的经验，keep 的性能非常好，因此如果它在大数据上具有最佳性能，我不会感到惊讶。

【讨论】：

我不明白。为什么会缺少城市名称？

@BarbarosÖzhan 。 . .谢谢你。我确定了答案。

【参考方案3】：

感谢您的回复。我会检查“保留”一次。 我是这样解决的。有没有其他的可能来优化它或者让它更简单？

**

select city,length(city) as "SL"

from (select city from station  order by city asc) 

where length(city) in (select min(length(city)) from station) and rownum=1; 

select city,length(city) as "LL"

from (select city from station  order by city asc) 

where length(city) in (select max(length(city)) from station) 

order by city desc;`

**

【参考方案4】：

对于 MS SQL Server，解决方案如下。

Declare @Small int
Declare @Large int
select @Small = Min(Len(CITY )) from STATION 
select @Large = Max(Len(CITY )) from STATION
;with cte1 as(
select Top 1 CITY  as SmallestCity,Len(CITY ) as Minimumlength from STATION  where Len(CITY ) = @Small Order by CITY  Asc
),
cte2 as(
select Top 1 CITY  as LargestCity,Len(CITY ) as MaximumLength from STATION  where Len(CITY ) = @Large Order by CITY  Asc 
)
select * from cte1
union
select * from cte2

它的输出看起来像这样， Out put of query

【参考方案5】：

我编写的代码适用于 Microsoft SQL Server

select city, len(city) from station where len(city) = (select max(len(city)) from station); 

select city, len(city) from station where len(city) = (select min(len(city)) from station) order by city desc offset 2 rows

【讨论】：

正如目前所写，您的答案尚不清楚。请edit 添加其他详细信息，以帮助其他人了解这如何解决所提出的问题。你可以找到更多关于如何写好答案的信息in the help center。

【参考方案6】：

你需要修改你的ORDER BY:

SELECT CITY, LENGTH(CITY)
FROM STATION
ORDER BY LENGTH(CITY) DESC, CITY
FETCH FIRST 1 ROWS ONLY ;

这将给出最长名字的答案。对于最短的名称，只需将ORDER BY 更改为使用ASC 为LENGTH(CITY)。

【讨论】：

要求要求两个城市名称，而不仅仅是一个。

我确实阅读了该部分，但用户没有指定它必须在一个查询中。不过，您的回答会解决这个问题。

【参考方案7】：

SELECT CITY,
       LENGTH(CITY) as length
  FROM CITY
 ORDER BY length ASC limit 1;

【讨论】：

limit 选项 ORDER 子句在 Oracle 中不存在。

虽然这个答案可能是正确的。仅代码答案很少有帮助。请评论您的代码并解释此代码如何解决问题。

【参考方案8】：

选择城市，长度（城市）

从车站

WHERE LENGTH(CITY) IN (SELECT min(LENGTH(CITY))

                                   FROM  STATION
                            )

AND ROWNUM

联合

选择城市，长度（城市）

从车站 WHERE LENGTH(CITY) IN (SELECT max(LENGTH(CITY))

                               FROM  STATION )

AND ROWNUM

按城市 ASC 排序；

【参考方案9】：

SELECT top 1 City,LEN(City) as Citylength into #tmp
FROM STATION
WHERE LEN(City)=(SELECT MIN(LEN(City)) FROM STATION )
order by City asc

SELECT top 1 City,LEN(City) as Citylength into #tmp1
FROM STATION
WHERE LEN(City)=(SELECT MAX(LEN(City)) FROM STATION )
order by City

select * from #tmp
union
select * from #tmp1