如何編寫一個函數來計算字符串中的連續字母相同

Question

searchingWrite函數conseclets它將接收一個字符串作爲參數。該函數將確定字符串中兩個或多個連續字母相同的所有情況。

例如，如果「Barrymoore」發送到該函數，則會說該字符串中有連續的字母r和o。但是如果「布什」被髮送到函數，它會說沒有兩個連續的字母是相同的。

這裏是我的代碼與它的問題是，當我把在信中找到它找到它但不連續

#include <iostream> 
#include <string> 
using namespace std; 

int main() 

{ 
    char searching='\0'; 
    string name=" "; 
    int counter =0; 

    cout<<"Enter a name : "<<endl; 
    getline(cin, name); 
    cout<<"Which letter would you like to count the number of times it appears: "<<endl; 
    cin>>name; 
    for(int i=0; i<name.length();i++){ 
     if(sentence[i]==searching){ 

     counter++; 

     } 
    } 
    cout<<"The letter " << searching << " appears "<< counter << " times "; 

    return 0; 


} 

Answer 1

#include <stdio.h> 
#include <stdlib.h> 
#include<string.h> 


int main(int argc, char *argv[]) { 
char a[30]; 
int i,c=0; 
printf("enter a string\n"); 
gets(a); 
for(i=0;i<strlen(a);i++) 
{ 
if(a[i]==a[i+1]) 
{ 
    printf("%c is consecutive\n",a[i]); 
    c++; 
} 
} 
if(c==0) 
{ 
printf("No consecutive letters"); 
} 
return 0; 
} 
// Check this. This is proper code for your problem. 

Answer 2

這裏是利用標準算法庫的實現。我稱之爲「運行」了相同連續字母的序列。該算法不一定是最佳的，但它很簡單，也很重要。

void conseclets(std::string const &str) { 
    // Keep the end of the string, and point i to the first run's beginning 
    auto e = end(str), i = std::adjacent_find(begin(str), e); 

    if(i == e) 
     std::cout << "No repetition.\n"; 
    else do { 
     // Locate the end of the run (that is, the first different letter) 
     auto next = std::find_if(i, e, [&i](auto const &c){ return c != *i; }); 

     // Print out the match 
     std::cout << "Letter " << *i << " is repeated " 
        << std::distance(i, next) << " times.\n"; 

     // Skip to the next run's beginning 
     i = std::adjacent_find(next, e); 

    // Do so until we reached the end of the string 
    } while(i != e); 
} 

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Answer 3

#include <iostream> 
#include <iterator> 
using namespace std; 

template<typename I, typename O> O repeats(I b, I e, O result){ 
    while(b!=e){ 
     bool once = true; 
     I p = b; 
     while(++b!=e && *p==*b){ 
      if(once){ 
       *result++ = *p; 
       once = false; 
      } 
     } 
    } 
    return result; 
} 

int main() {  
    string name = "Barrymoore"; 
    string res; 
    repeats(begin(name), end(name), back_inserter(res)); 
    cout << "There are " << res.size() << " consecutive letters :" << res << endl; 
    return 0; 
} 

Answer 4

雖然有幾種方法可以做到這一點，只是修改你的最佳實現什麼要求在這裏：

#include <iostream> 
using namespace std; 

int main() { 
    cout << "Hello, World!" << endl; 
    char searching = '\0'; 
    string name = " "; 
    int counter = 0; 
    cout << "Enter a name : " << endl; 
    getline(cin, name); 
    cout << "Which letter would you like to count the number of times it appears: " << endl; 
    cin >> searching; 
    for (int i = 0; i < name.length(); i++) { 
     if (name[i] == searching) { 
      counter++; 
     } 
    } 
    cout << "The letter " << searching << " appears " << counter << " times "; 
    return 0; 
}