Question

我正在制作一个程序，以基本上显示有关用户输入的单词的统计信息。到目前为止，该程序的其余部分都还不错，但是我很难为类型vector的{{1}}添加单词。

我环顾四周，找到了几个答案，我以为可以解决我的问题，但是我遇到了一个非常奇怪的编译器错误，或者它根本不起作用。我尝试将WordCount和emplace_back用于我认为正确的呼叫。本质上，我的问题代码如下：

push_back

为此，我必须使用向量，并且不能按照指示使用指针（如我所见）或#include <iostream> #include <string> #include <vector> using namespace std; //for simplicity here struct WordCount { string word; int count; //I have tried using this too: WordCount(string _word, int _count) : word{_word}, count{_count} {} }; //...// void wordToVector(/**...**/,string addStr, vector<WordCount>& wordStats){ /**... code that I've tested to work; basically determined if the word was already said as I need to have unique words only...**/ wordStats.push_back(WordCount(addStr, 1)); /** also tried: (some had "#include <istream>" when using emplace_back but that didn't seem to make a difference for me in any case) wordStats.emplace_back(WordCount(addStr, 1)); wordStats.emplace_back({addStr, 1}); wordStats.push_back(addStr, 1) wordStats.push_back(addStr).word; (and wordStats.push_back(1).count;) **/ } int main() { vector<WordCount> wordStats(1); //"1" to initialize the size wordStats.at(0).word = ""; wordStats.at(0).count = 0; /**There's already a part to change the first values to what they should be, and it worked last I tested it. Below is a part was for my personal use to see if anything came out... if it worked**/ for (int i = 0; i < 3; i++) { cout << wordStats.at(i).word << endl; cout << wordStats.at(i).count << endl; } return 0; }。如果我输入“ Oh happy day！”，它应该可以打印（固定后，使用当前的cout语句）：

OH

1

快乐

1

DAY

1

（有一个较早的部分将每个字母都大写，我测试过可以使用）。这是我在这里的第一篇文章，因为我迷路了。如果我提供过多或不足，请让我知道。 **编辑格式

Answer 1

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

struct WordCount {
    string word;
    int count;
};

void wordToVector(string addStr, vector<WordCount>& wordStats){
    for (int i = 0; i < wordStats.size(); i++) {
        if (wordStats[i].word == addStr) {
            wordStats[i].count = wordStats[i].count + 1;
            return;
        }
    }
    struct WordCount wc;
    wc.word = addStr;
    wc.count = 1;
    wordStats.push_back(wc);
}

int main() {
    vector<WordCount> wordStats;
    wordToVector("hehe", wordStats);
    wordToVector("hehe", wordStats);
    wordToVector("haha", wordStats);


    for (int i = 0; i < wordStats.size(); i++) {
        cout << wordStats.at(i).word << endl;
        cout << wordStats.at(i).count << endl;
    }
    return 0;
}

使用此代码，我得到输出：

hehe
2
haha
1

还有什么需要补充的吗？

如果要用空格分割输入并检查输入中每个单词的出现，则较长的文本检查每个单词可能效率很低（我想这是线性的，M * N复杂度）如果允许的话，我建议您使用以单词为键，值作为出现次数的地图-或类似的方式。

尝试向使用结构

1 个答案: