The task of finding the shortest word in the text.

Get the string.

Set up four variables:

1. Counter - the length of the current word;
2. Position of the beginning of the current word;
3. The length of the shortest word encountered;
4. Position of the beginning of the shortest word encountered.

We start in the loop to pass the string character by character. For fixed-length character encodings (a la CP1251, CP866 or ISO-8859-1), everything is simple; for variable-length character encodings (for example, UTF-8), you need to be more cunning. Shkolote and non-core and / or young students, as a rule, do not talk about these difficulties, so if there is no understanding of what it is about, then you can ignore the paragraph. Until.

If the next symbol is a letter (a letter or another, a symbol included in the “word”), then:

• If the current word length counter was 0 (no current word), then remember the position.
• Make +1 to the length of the current word.

If not litera:

• If the length of the current word is greater than 0 (i.e. there was something), then compare it with the length of the shortest word encountered.
• Found shorter? Remember the length and position.
• We reset the length of the current word to 0.

So we drive to the end of the line, or until the word is found in 1 character (i.e. there is no further search for it).

Next thing is the technique - the position in the string we know, the length of the word (or the fact that there were no words in the string) - we know.

No arrays and other circus is needed here. It is possible, but not necessary. Dumb state machine, search in one pass, for O(n) .

Here is the whole basic idea. Code - please write yourself, in the end this is your task. If something is not clear, ask specific questions (and how you tried to solve them).

The algorithm is about the same (very simple, you can certainly optimize):

1. Assign the sentence to a string variable.
   
2. We break the sentence into words and store it in an array. For example, by space, comma, etc.
   
3. We get a variable and assign it the length of the first word ( minLenght , for example). We also set a variable to store the index.
   
4. Then in the cycle we run through the array of words and compare the length of the word with the variable that we have ( minLength ). If the length is less, then assign this variable the length of the current word.
   
5. Print the smallest word, know the length and know the index in the array.

I, of course, in Delphi is not strong, but these are the basics! Want to achieve a result - memorize the basics. The syntax of the language is not so important. The point is this: it is impossible to predict in advance what the user will enter. He can create a string with just one word. And maybe four to try to dial. A string is, in any case, an array of characters. Consider the concept of "word" in the context of the compiler: the word is a sequence of characters, separated by spaces. It is us, people, who take words as something meaningful, which allows us to express our thoughts. However, for computer technology it is not. I will give an example: рпавол вфушщо мтвлопопра . If you take roughly - it is a line of three words and it does not matter that they do not make sense.

Well, here we have an understanding of the essence of the problem. We make two counters: one to determine the continuous sequence of characters (the length of the word differently), the other to memorize the number of the longest word. That's all. We take an array of these characters (the string entered by the user) and loop. Count the number of characters other than a space, meet a space - save the number in an array variable that stores word lengths, etc. At the output, we obtain an array in which the lengths of the words are located, of which we must find the maximum and display it on the screen. Well, with the conclusion of the number of the longest word, I think, there will be no problems.

Here is the solution to this problem using OOP:

 program Project3; {$APPTYPE CONSOLE} uses Classes; type TLst = class(TStringList) function CompareStrings(const a,b: string): integer; override; end; function TLst.CompareStrings; begin Result := Length(a)-Length(b); if Result = 0 then Result := inherited CompareStrings(a,b) end; var s : string; begin with TLst.Create do try Sorted := true; repeat ReadLn(s); DelimitedText := s; if Count > 0 then WriteLn('<',Strings[0],'>') until s = '' finally Free end end. 

Moreover, an arbitrary number of lines is processed (in the loop repeat ReadLn (s) until the empty line is entered - until s = ''). Thus, the main body of the Repeat loop consists of only two operators:

 DelimitedText := s; // - очередную строку разбить на слова и отсортировать WriteLn(Strings[0]); // - верхний элемент получившегося списка и есть ответ 

Where the leading statement is if if Count> 0 then ... is "protection" from an empty list.

The fact that the split list of words will be sortable is determined by assigning the Sorted: = true property, and the correct sorting for our task is established by redefining the CompareStrings function (in accordance with the spirit and the OOP letter) in the heir of TStringList , which we used as the basis:

"... A robe is the same, but the buttons are nacreous ..." (c) c / f The diamond hand.

It should be noted that the sorting set here is used even more sophisticated than indicated in the original task, since it does not indicate what to do if the initial line contains several words with a minimum length - who should I choose? Therefore, there is a sorting according to a double feature: each comparison goes first along the length of the words, and then, with an equal length, alphabetically, i.e. traditional lexicographical order is used (due to the use of inherited ... ).

If desired, if it is necessary to distinguish between upper and lower case letters (by default, they do not differ here), simply assign the corresponding value to the CaseSensitive property.

I would especially like to note that here an anonymous list is used (developed by us when inheriting TStringList - TLst ) without using a variable of this type in the var section and therefore (using the with operator) all statements inside a protected try-finally-end block are short and descriptive , and the program itself is understandable, transparent and easily modified.