C++统计vector字符串出现次数：越界与最后元素未统计问题

自学C++一周，编写了统计vector<string>容器words中各字符串出现次数的代码。当前代码可正常输出「Six」「Three」「Two」的出现次数，但无法输出「One」的统计结果，且程序会因访问vector越界报错。需要解决：在不越界的前提下判断当前元素与下一个元素是否相等，同时确保最后一个元素被正确统计。

原代码如下：

int main()
{
    vector<string> words;
    words.push_back("Six");
    words.push_back("Six");
    words.push_back("Six");
    words.push_back("Six");
    words.push_back("Six");
    words.push_back("Six");
    words.push_back("Three");
    words.push_back("Three");
    words.push_back("Three");
    words.push_back("Two"); 
    words.push_back("Two");
    words.push_back("One");

    vector<int> values; // vector to tracks the # of times a string
                        //appears in the vecotor words

    //create elements in values equal to number of 
    //elements in words
    //note: this creates too many elements
    //I am aware but am a begginner and don't know how
    //to solve this issue
    for (int i = 0; i <= words.size() - 1; i++) 
    {
        values.push_back(0);
    }

    int invariant = 0; // tracks number of times loops
    string current; // used make sure duplicate strings aren't recounted

    while (invariant <= words.size() - 1) 
    {
        //compare each element of words to each other element 
        //one at a time and increment the appropriate
        //element of values if they match
        for (int i = 0; i <= words.size() - 1; i++)
        {
            if (words[invariant] == words[i])
            {
                values[invariant] = values[invariant] + 1;
            }
        }
        //used to check for duplicate strings
        current.clear();
        current = words[invariant];

        if (invariant + 1 > words.size()) //avoids going outside range of words
        {
            return 0; //Errors outside range after this return
        }
        else{
            if (words[invariant+1] == current) // if the next word is the same
            {
                ++invariant;
            }
            else //if the next word is different output the number of times the current word appears
            {
                cout << current << " appears " << values[invariant] << " times.\n";
                ++invariant;
            }
        }
    }
    return 0;
}

注：习惯在循环中使用invariant <= value的写法，若不设置invariant == value - 1程序就会出现越界错误。

核心问题 #

• 越界判断逻辑错误：原代码中invariant + 1 > words.size()的判断无效，当invariant指向最后一个元素（索引为words.size()-1）时，invariant+1等于words.size()，此时访问words[invariant+1]会直接触发越界。
• 最后一个元素未输出：处理到最后一个元素时，代码直接return 0，导致「One」的统计结果无法输出。
• 冗余的values容器：创建了与words等长的values容器，实际上只需记录当前字符串的出现次数即可。

修正后的代码 #

#include <iostream>
#include <vector>
#include <string>

using namespace std;

int main()
{
    vector<string> words;
    words.push_back("Six");
    words.push_back("Six");
    words.push_back("Six");
    words.push_back("Six");
    words.push_back("Six");
    words.push_back("Six");
    words.push_back("Three");
    words.push_back("Three");
    words.push_back("Three");
    words.push_back("Two"); 
    words.push_back("Two");
    words.push_back("One");

    vector<int> values;
    for (int i = 0; i <= words.size() - 1; i++) 
    {
        values.push_back(0);
    }

    int invariant = 0;
    string current;

    while (invariant <= words.size() - 1) 
    {
        // 统计当前字符串出现次数
        for (int i = 0; i <= words.size() - 1; i++)
        {
            if (words[invariant] == words[i])
            {
                values[invariant]++;
            }
        }
        current = words[invariant];

        // 优先判断是否为最后一个元素，避免越界
        if (invariant == words.size() - 1)
        {
            cout << current << " appears " << values[invariant] << " times.\n";
            ++invariant;
        }
        else
        {
            if (words[invariant+1] == current)
            {
                ++invariant;
            }
            else
            {
                cout << current << " appears " << values[invariant] << " times.\n";
                ++invariant;
            }
        }
    }
    return 0;
}

关键修改点 #

• 修正越界问题：新增invariant == words.size() - 1的判断，直接处理最后一个元素的输出，彻底避免访问words[invariant+1]导致的越界。
• 保留你的循环习惯：依然使用invariant <= words.size() - 1的循环条件，符合你的编程习惯。
• 确保最后一个元素输出：当invariant指向最后一个元素时，直接输出统计结果，不再提前return。

进阶优化（可选） #

如果想简化代码，无需手动处理重复和越界问题，可以用std::map自动统计次数，代码更简洁高效：

#include <iostream>
#include <vector>
#include <string>
#include <map>

using namespace std;

int main()
{
    vector<string> words = {"Six", "Six", "Six", "Six", "Six", "Six", "Three", "Three", "Three", "Two", "Two", "One"};
    
    map<string, int> countMap;
    for (const string& word : words)
    {
        countMap[word]++;
    }

    for (const auto& pair : countMap)
    {
        cout << pair.first << " appears " << pair.second << " times.\n";
    }

    return 0;
}

内容的提问来源于stack exchange，提问作者MetalGooseSolid