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;
}
答案 0 :(得分:7)
是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;
}