C ++：如何检测向量中的重复项< string>并打印一份？ [英] C++ : How to detect duplicates in vector<string> and print ONE copy?

问题描述

我是C ++的新手。我想知道如何在一个向量中找到重复的字符串，并打印出一个字符串的副本。例如，如果我有<猫，狗，狗，鸟，>它会打印出猫，狗，鸟。我已经排序我的向量和我使用adjacent_find函数和迭代通过向量（因为我必须找到，如果任何单词是重复的）。我的代码检测重复，但它只打印非重复。我想改变它打印出所有的非重复，也只有一个重复，所以在矢量中的所有字符串都打印出来。这是我到目前为止的代码：

I'm new to C++. I was wondering how I can find duplicate strings in a vector and print out ONE copy of the string. For example, if I had <"cat", "dog", "dog", "bird",> it would print out cat, dog, bird. I have sorted my vector and am using the adjacent_find function and iterating through the vector (since I have to find if any word is duplicated). My code detects duplicates, but it only prints out the non-duplicates. I would like to alter it to print out all the non-duplicates and also just ONE of the duplicates, so all strings in the vector are printed out. Here is the code I have so far:

public: void print(vector<string> in) // print method for printing a vector and it's key
{ 

  sort(in.begin(), in.end()); // sort the vector alphabetically first

  vector<string>::iterator it; 

      for( it = in.begin(); it != in.end(); it++ ) // iterate through it


             if(adjacent_find(in.begin(), in.end()) == in.end()) // don't print duplicates


             cout << *it<<endl; // and print out each string in the vector
}

推荐答案

您可以使用STL算法 std :: unique（）或 std :: unique_copy（）。

You can use the STL algorithms std::unique() or std::unique_copy(). They work with any STL container, not just vectors.

将向量打印到标准输出的一个简单示例：

A simple example to print the vector to the standard output:

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

using namespace std;

int main()
{
    vector<string> v = { "hello", "hello", "world" };
    unique_copy(begin(v), end(v), ostream_iterator<string>(cout, " "));
}

如果要现场执行此操作， code> std :: unique（）。重要的是要记住，此函数不会物理删除多余的元素，但它返回迭代器到集合的新逻辑结束：

If you want to perform this operation in-place instead, you can use std::unique(). It is important to keep in mind that this function does not physically remove the redundant elements, but it returns the iterator to the new logical end of the collection:

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

using namespace std;

int main()
{
    vector<string> v = { "hello", "hello", "world" };
    auto newEnd = unique(begin(v), end(v));
    for_each(begin(v), newEnd, [] (string const& s) { cout << s << " "; });
}

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