C++中的字符串反转

你的方法很好。但你也可以这样做:

• 空间
• S
• 如果您发现空间,请将 Q

N 书堆上的字 N-1 队列中的数字。

While stack not empty do
 print S.pop
 if stack is empty break
 print Q.deque number of spaces
end-while

这里有一个方法。

#include <iostream>
#include <algorithm>
#include <vector>
#include <string>
#include <sstream>
using namespace std;

string test_string = "this is my test    string";

int main()
{
    // Create 2 vectors of strings.  One for words, another for spaces.
    typedef vector<string> strings;
    strings words, spaces;
    // Walk through the input string, and find individual tokens.
    // A token is either a word or a contigious string of spaces.
    for( string::size_type pos = 0; pos != string::npos; )
    {
        // is this a word token or a space token?
        bool is_char = test_string[pos] != ' ';
        string::size_type pos_end_token = string::npos;

        // find the one-past-the-end index for the end of this token
        if( is_char )
            pos_end_token = test_string.find(' ', pos);
        else
            pos_end_token = test_string.find_first_not_of(' ', pos);

        // pull out this token
        string token = test_string.substr(pos, pos_end_token == string::npos ? string::npos : pos_end_token-pos);
        // if the token is a word, save it to the list of words.
        // if it's a space, save it to the list of spaces
        if( is_char )
            words.push_back(token);
        else
            spaces.push_back(token);
        // move on to the next token
        pos = pos_end_token;
    }

    // construct the new string using stringstream
    stringstream ss;
    // walk through both the list of spaces and the list of words,
    // keeping in mind that there may be more words than spaces, or vice versa
    // construct the new string by first copying the word, then the spaces
    strings::const_reverse_iterator it_w = words.rbegin();
    strings::const_iterator it_s = spaces.begin();
    while( it_w != words.rend() || it_s != spaces.end() )
    {
        if( it_w != words.rend() )
            ss << *it_w++;
        if( it_s != spaces.end() )
            ss << *it_s++;
    }

    // pull a `string` out of the results & dump it
    string reversed = ss.str();
    cout << "Input: '" << test_string << "'" << endl << "Output: '" << reversed << "'" << endl;

}

我可以这样重新表述这个问题:

• 非空格标记被反转,但保留其原始顺序
  • 这5个非空格标记this,is,my,test,string被反转为string,test,my,is,this。
• 太空代币保持原来的顺序

下面是一个O(N)解[N是char数组的长度]。不幸的是,它并没有按照OP的要求就位,但是它也没有使用额外的堆栈或队列——它使用一个单独的字符数组作为工作空间。

这是一个C-ish伪代码。

work_array = char array with size of input_array
dst = &work_array[ 0 ]

for( i = 1; ; i++) {
   detect i’th non-space token in input_array starting from the back side
   if no such token {
      break;
   }
   copy the token starting at dst
   advance dst by token_size
   detect i’th space-token in input_array starting from the front side
   copy the token starting at dst
   advance dst by token_size
}

// at this point work_array contains the desired output,
// it can be copied back to input_array and destroyed

对于从第一个词到中心词的单词,用单词长度-n切换单词n

此伪代码假定初始字符串不以空格结尾,但也可以对此进行适当修改。

1.  Get string length; allocate equivalent space for final string; set getText=1

2.  While pointer doesn't reach position 0 of string,

    i.start from end of string, read character by character... 
      a.if getText=1 
       ...until blank space encountered
      b.if getText=0
       ...until not blank space encountered

    ii.back up pointer to previously pointed character

    iii.output to final string in reverse

    iv.toggle getText

3.  Stop

并非所有strtok解决方案都适用于您的示例,请参见上文。 试试这个:

char *wordrev(char *s)
{
  char *y=calloc(1,strlen(s)+1);
  char *p=s+strlen(s);
  while( p--!=s )
    if( *p==32 )
      strcat(y,p+1),strcat(y," "),*p=0;
  strcpy(s,y);
  free(y);
  return s;
}

#include <string>
#include <iostream>
#include <algorithm>

class push_front
{
public:
   push_front( std::string& s ) : _s(s) {};
   bool operator()(char c) { _s.insert( _s.begin(), c ); return true; };
   std::string& _s;
};

int main( int argc, char** argv )
{

   std::string s1;
   std::string s( "Now is the time for all good men");
   for_each( s.begin(), s.end(), push_front(s1) );

   std::cout << s << "\n";
   std::cout << s1 << "\n";
}

不,不,不,不,不

复制数组中的每个字符串并按相反顺序打印(i--)

int main()
{
int j=0;
string str;
string copy[80];
int start=0;
int end=0;
cout<<"Enter the String :: ";
getline(cin,str);
cout<<"Entered String is : "<<str<<endl;
for(int i=0;str[i]!='\0';i++)
{
end=s.find(" ",start);
if(end==-1)
{
copy[j]=str.substr(start,(str.length()-start));
break;
}
else
{
copy[j]=str.substr(start,(end-start));
start=end+1;
j++;
i=end;
}
}

for(int s1=j;s1>=0;s1--)
cout<<" "<<copy[s1];
}

我想我应该使用空格字符对字符串进行标记化(strtok或CString::Tokanize)。将字符串放入一个向量中,然后按相反的顺序将它们拉回来,并在它们之间用空格连接起来。