如何将多个字符串向量组合成一个向量？

import itertools

# 假设有多个字符串向量
vector1 = ['a', 'b', 'c']
vector2 = ['d', 'e', 'f']
vector3 = ['g', 'h', 'i']

# 使用 itertools.chain
combined_vector = list(itertools.chain(vector1, vector2, vector3))
print(combined_vector)  # 输出: ['a', 'b', 'c', 'd', 'e', 'f', 'g', 'h', 'i']

# 或者使用列表推导式
combined_vector = vector1 + vector2 + vector3
print(combined_vector)  # 输出: ['a', 'b', 'c', 'd', 'e', 'f', 'g', 'h', 'i']

// 假设有多个字符串向量
let vector1 = ['a', 'b', 'c'];
let vector2 = ['d', 'e', 'f'];
let vector3 = ['g', 'h', 'i'];

// 使用 concat 方法
let combinedVector = vector1.concat(vector2, vector3);
console.log(combinedVector);  // 输出: ['a', 'b', 'c', 'd', 'e', 'f', 'g', 'h', 'i']

import java.util.ArrayList;
import java.util.Arrays;
import java.util.List;

public class Main {
    public static void main(String[] args) {
        // 假设有多个字符串向量
        List<String> vector1 = Arrays.asList("a", "b", "c");
        List<String> vector2 = Arrays.asList("d", "e", "f");
        List<String> vector3 = Arrays.asList("g", "h", "i");

        // 创建一个新的 ArrayList 来存储合并后的结果
        List<String> combinedVector = new ArrayList<>();
        combinedVector.addAll(vector1);
        combinedVector.addAll(vector2);
        combinedVector.addAll(vector3);

        System.out.println(combinedVector);  // 输出: [a, b, c, d, e, f, g, h, i]
    }
}

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

int main() {
    // 假设有多个字符串向量
    std::vector<std::string> vector1 = {"a", "b", "c"};
    std::vector<std::string> vector2 = {"d", "e", "f"};
    std::vector<std::string> vector3 = {"g", "h", "i"};

    // 创建一个新的 vector 来存储合并后的结果
    std::vector<std::string> combinedVector;
    combinedVector.insert(combinedVector.end(), vector1.begin(), vector1.end());
    combinedVector.insert(combinedVector.end(), vector2.begin(), vector2.end());
    combinedVector.insert(combinedVector.end(), vector3.begin(), vector3.end());

    for (const auto& str : combinedVector) {
        std::cout << str << " ";
    }
    std::cout << std::endl;  // 输出: a b c d e f g h i

    return 0;
}