输入和输出迭代器

Question

我有一个关于使用输入和输出迭代器的练习题。功能的标题是如下

template<class InpIter,class OutpIter> 
OutpIter my_unique_copy(InpIter first, InpIter last, OutpIter result) 

该函数应的元素从所述范围复制[第一，最后]导致。在连续的重复元素组中，只有第一个值被复制。返回值是元素复制到的范围的末尾。 复杂性：线性

我有一个想法，做刚琢磨了一点帮助，因为我不是迭代器舒适又不失

template<class InpIter,class OutpIter> 
OutpIter my_unique_copy(InpIter first, InpIter last, OutpIter result){ 
    InpIter current=first; 
    first++;//point to second element 
    while(first!=last){ 
     while(*first==*current){//Keep comparing elements to current to see if they're same 
      first++; 
     } 
     result=current; 
     current=first; 
     result++; 
     first++; 
    } 
    return result; 
} 

Answer 1

我想这是你所追求的，什么每一步的解释。

template<class InpIter,class OutpIter> 
OutpIter my_unique_copy(InpIter first, InpIter last, OutpIter result) 
{ 
    // keep going until end of sequence 
    while(first!=last) 
    { 
     // save current position. 
     InpIter current=first; 

     // advance first, test it against last, and if not 
     // equal, test what it references against current. 
     // repeat this until *first != *current, or first == last 
     while (++first != last && *first == *current); 

     // not matter what dropped the loop above, we still have 
     // our current, so save it off and advance result. 
     *result++ = *current; 
    } 

    // not sure why you want to return the first iterator position 
    // *past* the last insertion, but I kept this as-is regardless. 
    return result; 
} 

我希望能解释一下。 （我不认为我错过了什么，但我敢肯定，我会听到它，如果我做到了。）

一个非常简单的测试工具：

#include <iostream> 
#include <iterator> 
#include <algorithm> 

int main() 
{ 
    int ar[] = { 1,2,2,3,3,3,4,5,5,6,7,7,7,7,8 }; 
    int res[10] = {0}; 

    int *p = my_unique_copy(std::begin(ar), std::end(ar), res); 
    std::copy(res, p, std::ostream_iterator<int>(std::cout, " ")); 
    return 0; 
} 

输出

1 2 3 4 5 6 7 8