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Question

我想写一个函数，它将返回无重复数字的数量。到目前为止，我已经编写了一个函数，它迭代通过char并收集到所有数字的向量，但问题来自于我必须从向量仅获得非重复数字。

我的代码：

int norepeat(char *word){
    int i = 0;
    int size = 0;
    vector<int> tab;
    while (word[i] != '\0'){
        if (word[i] >= '0' && word[i] <= '9') {
            int num = word[i];
            tab.push_back(num);
            sort(tab.begin(),tab.end());
            unique(tab.begin(),tab.end());
            size = tab.size();
        }
        ++i;
    }
    return size;
}

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几个例子应该如何运作：

norepeat("de32ge2sa3ds1") => 1
norepeat("defegtdsg") => 0
norepeat("12341234") => 0
norepeat("1yle2le49") => 4

Answer 1

创建数字地图 - ＆gt;数字计数。

然后遍历地图并计算计数为1的位数。

int norepeat(char *word){
   int i = 0;
   std::map<char, int> m;
   while (word[i] != '\0'){
      if ( isdigit(word[i] )
         m[word[i]]++;
      ++i;
   }

   int count = 0;
   for ( auto& p : m )
   {
      if ( p.second == 1 )
         ++count;
   }

   return count;
}

使用不支持C ++ 11的编译器时，for循环可以更改为：

   std::map<char, int>::iterator iter = m.begin();
   std::map<char, int>::iterator end = m.end();
   for ( ; iter != end; ++iter )
   {
      if ( iter->second == 1 )
         ++count;
   }

Answer 2

C ++的优点是能够将来自C ++标准模板库的Plain-Old-Data指针和libc函数与算法混合使用:(注意：使用一些C ++ 11特性）

#include <vector>
#include <iostream>
#include <algorithm>
#include <iterator>
#include <vector>
#include <ctype.h>

using namespace std;

size_t norepeat(const char* word) {
    vector<char>            digits;
    vector<char>::iterator  uniq;

    // Filter the digits
    copy_if(word, word+strlen(word), back_inserter(digits), ::isdigit);

    // get the unique ones
    sort(digits.begin(), digits.end());
    uniq = unique(digits.begin(), digits.end());

    // return amount
    size_t uniques    = std::distance(digits.begin(), uniq);
    size_t duplicates = std::distance(uniq, digits.end());
    return uniques - duplicates;
}

int main( void ) {
    cout << norepeat("hello 123 world 124") << endl;
    cout << norepeat("hello world") << endl;
    return 0;
}

输出：

2
0

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只是为了笑：写了一个count-output-iterator-class;它会将复制到其中的值的出现次数计入value - >的地图中。 unsigned int。结合count_if算法（使用C ++ lambda），norepeat函数本身只有三个语句：变量声明，过滤数字并返回计数结果：

#include <map> #include <vector> #include <iostream> #include <algorithm> #include <iterator> #include <ctype.h> using namespace std; template <typename T> struct counter : public iterator<output_iterator_tag, T> { typedef map<T, unsigned int> counter_type; counter(counter_type* ptr) : mapptr( ptr ) {} // decorate with dereference and assignment counter<T>& operator=( const T& t ) { mapptr->operator[]( t )++; return *this; } counter<T>& operator++() { return *this; } counter<T>& operator*() { return *this; } counter_type* mapptr; }; size_t norepeat(const char* word) { typename counter<char>::counter_type countert; // Filter the digits copy_if(word, word+strlen(word), counter<char>(&countert), ::isdigit); // Count the ones that have a value of one return count_if(countert.begin(), countert.end(), [](const counter<char>::counter_type::value_type& kv) { return kv.second==1; } ); } int main( void ) { cout << norepeat("hello 123 world 124") << endl; cout << norepeat("hello world") << endl; return 0; }

Answer 3

由于我的上一个答案是基于对该问题的误读，这是怎么回事？

int norepeat(const char *word){
    int i = 0;
    int size = 0;
    int arr[10] = {0}; // there are only 10 unique possibilities,
                       // so we'll mark them when found
    while (word[i] != '\0')
    {
        if (std::isdigit(word[i])) 
        {
            int num = word[i] - '0'; // get numeric value of digit to use as index
            arr[num]++; // count the number of times we've seen this digit
        }
        ++i;
    }
    for (i = 0; i < 10; i++)
    {
        if (arr[i] == 1) // count all element seen only once
        {
            size++;
        }
    }
    return size;
}

这个插头是非常有限的，因为它只适用于十进制数字，但稍作修改就可以做大写或小写字符。有std::map ......可能性是无限的！我将在一秒钟内添加完整性。

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地图版。

int norepeat(const char *word){
    int i = 0;
    int size = 0;
    std::map<char, int> counts; 

    while (word[i] != '\0')
    {
        if (std::isdigit(word[i])) // could use any or no filtering logic here
        {
            counts[word[i]]++;
        }
        ++i;
    }
    for (auto &count: counts)
    {
        if (count.second == 1) // count all element seen only once
        {
            size++;
        }
    }
    return size;
}

Answer 4

int norepeat(char *word){

int i = 0;
int size = 0;
vector<int> tab;
while (word[i] != '\0'){
    if (word[i] >= '0' && word[i] <= '9') {
        int num = word[i];
        tab.push_back(num);
    }
    ++i;
}

size = std::distance( unique(tab.begin(),tab.end()), tab.end() );
return size;

}

矢量只获得重复元素

4 个答案:

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