在C＃中不使用String.Split反轉句子的單詞

Question

最近，有人在討論中要求我編寫一種算法，以實現句子中單詞的反向（不是整個句子的反向），而不使用諸如ToCharArray和Length之外的字符串操作，例如Split / Replace / Reverse / Join。 以下是我可以在5分鍾的時間內設計的內容。 盡管該算法運行良好，但實現方式似乎有點難看。 某些機構可以通過完善代碼來幫助我。

string ReverseWords(string s)
{
    string reverseString = string.Empty;
    string word = string.Empty;

    var chars = s.ToCharArray();
    List<ArrayList> words = new List<ArrayList>();
    ArrayList addedChars = new ArrayList();
    Char[] reversedChars = new Char[chars.Length];
    int i = 1;
    foreach (char c in chars)
    {
        if (c != ' ')
        {
            addedChars.Add(c);
        }
        else
        {
            words.Add(new ArrayList(addedChars));
            addedChars.Clear();
        }
        if (i == s.Length)
        {
            words.Add(new ArrayList(addedChars));
            addedChars.Clear();
        }
        i++;
    }
    foreach (ArrayList a in words)
    {
        for (int counter = a.Count - 1; counter >= 0; counter--)
        {
            reverseString += a[counter];
        }
        if(reverseString.Length < s.Length)
            reverseString += " ";
    }
    return reverseString;
}

Answer 1

有 一種使用LIFO堆棧相對簡潔的解決方案 。
但是，問題聽起來像是作業，所以我只提供偽代碼。

currWord = new LIFO stack of characters
while (! end of string/array)
{
  c = next character in string/array
  if (c == some_white_space_character) {
     while (currWord not empty) {
       c2 = currWord.pop()
       print(c2)
     }
     print(c)
  }
  else
    currWord.push(c)
}

Answer 2

這有點簡單：

string inp = "hai how are you?";
StringBuilder strb = new StringBuilder();
List<char> charlist = new List<char>();
for (int c = 0; c < inp.Length; c++ )
{

    if (inp[c] == ' ' || c == inp.Length - 1)
    {
        if (c == inp.Length - 1)
            charlist.Add(inp[c]);
        for (int i = charlist.Count - 1; i >= 0; i--)
            strb.Append(charlist[i]);

        strb.Append(' ');
        charlist = new List<char>();
    }
    else
        charlist.Add(inp[c]);
}
string output = strb.ToString();

Answer 3

拋光版本的種類：-

string words = "hi! how are you!";
string reversedWords = "";

List<int> spaceEncounter = new List<int>();
spaceEncounter.Add(words.Length - 1);

for (int i = words.Length - 1; i > 0; i--)
{ 
    if(words[i].Equals(' '))
    {
        spaceEncounter.Add(i);

        for (int j = i+1; j < spaceEncounter[spaceEncounter.Count - 2]; j++)
            reversedWords += words[j];

        reversedWords += " ";
    }
}

for (int i = 0; i < spaceEncounter[spaceEncounter.Count - 1]; i++)
    reversedWords += words[i];    

Answer 4

在C＃中使用堆棧

string str = "ABCDEFG";
Stack<char> stack=new Stack<char>();
foreach (var c in str)
{
    stack.Push(c);
}
char[] chars=new char[stack.Count];
for (int i = 0; i < chars.Length; i++)
{
    chars[i]=stack.Pop();
}
var result=new string(chars); //GFEDCBA

Answer 5

您的代碼中有一個小錯誤。 因此，輸出字符串將顯示為yo are how hi! ，給定輸入字符串hi! how are you hi! how are you 。 它會截斷最后一個單詞的最后一個字符。

更改此：

spaceEncounter.Add(words.Length - 1);

至：

spaceEncounter.Add(words.Length);

Answer 6

好吧，您沒有對其他LINQ擴展方法說什么：)

static string ReverseWordsWithoutSplit(string input)
{
    var n = 0;
    var words = input.GroupBy(curr => curr == ' ' ? n++ : n);

    return words.Reverse().Aggregate("", (total, curr) => total + string.Concat(curr.TakeWhile(c => c != ' ')) + ' ');
}

Answer 7

        string temp = string.Empty;
        string reversedString = string.Empty;

        foreach (var currentCharacter in testSentence)
        {
            if (currentCharacter != ' ')
            {
                temp = temp + currentCharacter;
            }
            else
            {
                reversedString = temp + " " + reversedString;
                temp = string.Empty;
            }
        }
        reversedString = temp + " " + reversedString;

Answer 8

此版本可就地工作，沒有任何中間數據結構。 首先，它反轉每個單詞中的字符。 “我也是” =>“ em oot”。 然后它反轉整個字符串：“ em oot” =>“ too me”。

    public static string ReverseWords(string s)
    {
        if (string.IsNullOrEmpty(s))
            return s;

        char[] chars = s.ToCharArray();
        int wordStartIndex = -1;

        for (int i = 0; i < chars.Length; i++)
        {
            if (!Char.IsWhiteSpace(chars[i]) && wordStartIndex < 0)
            {
                // Remember word start index
                wordStartIndex = i;
            }
            else
            if (wordStartIndex >= 0 && (i == chars.Length-1 || Char.IsWhiteSpace(chars[i + 1]))) {
                // End of word detected, reverse the chacacters in the word range
                ReverseRange(chars, wordStartIndex, i);

                // The current word is complete, reset the start index  
                wordStartIndex = -1;
            }
        }

        // Reverse all chars in the string
        ReverseRange(chars, 0, chars.Length - 1);

        return new string(chars);
    }

    // Helper
    private static void ReverseRange(char[] chars, int startIndex, int endIndex)
    {
        for(int i = 0; i <= (endIndex - startIndex) / 2; i++)
        {
            char tmp = chars[startIndex + i];
            chars[startIndex + i] = chars[endIndex - i];
            chars[endIndex - i] = tmp;
        }            
    }

Answer 9

一個簡單的遞歸函數如何檢查“”，然后相應地子字符串呢？

    private static string rev(string inSent) { 
        if(inSent.IndexOf(" ") != -1) 
        { 
            int space = inSent.IndexOf(" "); 
            System.Text.StringBuilder st = new System.Text.StringBuilder(inSent.Substring(space+1)); 
            return rev(st.ToString()) + " " + inSent.Substring(0, space); 
        } 
        else 
        { 
            return inSent; 
        } 
    }

Answer 10

最簡單的答案之一如下所示，請仔細閱讀，

public static string ReversewordString(string Name)
    {
        string output="";
        char[] str = Name.ToCharArray();
        for (int i = str.Length - 1; i >= 0; i--)
        {
            if (str[i] == ' ')
            {
                output = output + " ";
                for (int j = i + 1; j < str.Length; j++)
                {
                    if (str[j] == ' ')
                    {
                        break;
                    }
                    output=output+ str[j];
                }
            }
            if (i == 0)
            {
                output = output +" ";
                int k = 0;
                do
                {
                    output = output + str[k];
                    k++;
                } while (str[k] != ' ');
            }
        }
        return output;
    }