C++ 빅데이터 개발에서 데이터 병합 및 정렬 알고리즘을 최적화하는 방법은 무엇입니까?

#include <iostream>
#include <vector>
#include <thread>

// 归并排序的合并
void merge(std::vector<int>& arr, int left, int mid, int right) {
    int i = left;
    int j = mid + 1;
    int k = 0;
    std::vector<int> tmp(right - left + 1);  // 临时数组存放归并结果
    while (i <= mid && j <= right) {
        if (arr[i] <= arr[j]) {
            tmp[k++] = arr[i++];
        } else {
            tmp[k++] = arr[j++];
        }
    }
    while (i <= mid) {
        tmp[k++] = arr[i++];
    }
    while (j <= right) {
        tmp[k++] = arr[j++];
    }
    for (i = left, k = 0; i <= right; i++, k++) {
        arr[i] = tmp[k];
    }
}

// 归并排序的递归实现
void mergeSort(std::vector<int>& arr, int left, int right) {
    if (left < right) {
        int mid = (left + right) / 2;
        mergeSort(arr, left, mid);
        mergeSort(arr, mid + 1, right);
        merge(arr, left, mid, right);
    }
}

// 多线程排序的合并
void mergeThread(std::vector<int>& arr, int left, int mid, int right) {
    // 省略合并部分的代码
}

// 多线程归并排序的递归实现
void mergeSortThread(std::vector<int>& arr, int left, int right, int depth) {
    if (left < right) {
        if (depth > 0) {
            int mid = (left + right) / 2;
            std::thread t1(mergeSortThread, std::ref(arr), left, mid, depth - 1);
            std::thread t2(mergeSortThread, std::ref(arr), mid + 1, right, depth - 1);
            t1.join();
            t2.join();
            mergeThread(arr, left, mid, right);
        } else {
            mergeSort(arr, left, right);
        }
    }
}

int main() {
    std::vector<int> arr = {8, 4, 5, 7, 1, 3, 6, 2};
    
    // 串行排序
    mergeSort(arr, 0, arr.size() - 1);
    std::cout << "串行排序结果：";
    for (int i = 0; i < arr.size(); i++) {
        std::cout << arr[i] << " ";
    }
    std::cout << std::endl;

    // 多线程排序
    int depth = 2;
    mergeSortThread(arr, 0, arr.size() - 1, depth);
    std::cout << "多线程排序结果：";
    for (int i = 0; i < arr.size(); i++) {
        std::cout << arr[i] << " ";
    }
    std::cout << std::endl;

    return 0;
}