在C#中不使用String.Split反转句子的单词

时间:2012-12-24 06:30:54

标签: c# .net string algorithm

最近我在讨论中被要求写一个算法来实现一个句子的单词反转(不是整个句子的反转),而不使用除了ToCharArray和Length之外的字符串操作,如Split / Replace / Reverse / Join。以下是我在5分钟内设计的内容。虽然算法工作正常,但似乎有点丑陋的实现方式。有些人可以通过抛光代码来帮助我。

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;
}

10 个答案:

答案 0 :(得分:7)

一个相对优雅的解决方案,使用LIFO堆栈
然而问题听起来像家庭作业,所以我只提供伪代码。

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)
}

答案 1 :(得分:5)

这有点简单:

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();

答案 2 :(得分:2)

有点抛光版: -

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];    

答案 3 :(得分:2)

在C#中使用堆栈

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

答案 4 :(得分:1)

您的代码中存在一个小错误。因此,在输入字符串yo are how hi!的情况下,输出字符串将显示为hi! how are you。它正在截断最后一个字的最后一个字符。

改变这个:

spaceEncounter.Add(words.Length - 1);

要:

spaceEncounter.Add(words.Length);

答案 5 :(得分:1)

好吧,你没有说过其他LINQ扩展方法:)

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 != ' ')) + ' ');
}

答案 6 :(得分:0)

        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;

答案 7 :(得分:0)

此版本就地工作,没有任何中间数据结构。首先,它会反转每个单词中的字符。 &#34;我也是#34; =&GT; &#34; em oot&#34;。然后它反转整个字符串:&#34; em oot&#34; =&GT; &#34;我也是#34;。

    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;
        }            
    }

答案 8 :(得分:0)

检查&#34;的简单递归函数怎么样? &#34;然后相应地子串?

    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; 
        } 
    }

答案 9 :(得分:0)

最简单的答案之一如下所示,请仔细阅读,

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;
    }