以下是一个比较两个长列表并以最短的时间返回索引的解决方法的示例代码：

import multiprocessing

def compare_lists(list1, list2, start_index, end_index, result_queue):
    min_diff = float('inf')
    min_index = -1
    
    for i in range(start_index, end_index):
        diff = abs(list1[i] - list2[i])
        if diff < min_diff:
            min_diff = diff
            min_index = i
    
    result_queue.put((min_index, min_diff))

def compare_long_lists(list1, list2):
    num_processes = multiprocessing.cpu_count() # 获取CPU核心数
    chunk_size = len(list1) // num_processes # 划分每个进程处理的数据量
    
    processes = []
    result_queue = multiprocessing.Queue() # 创建一个进程间通信的队列
    
    # 创建并启动子进程
    for i in range(num_processes):
        start_index = i * chunk_size
        end_index = start_index + chunk_size
        
        # 最后一个进程处理剩余的数据
        if i == num_processes - 1:
            end_index = len(list1)
        
        p = multiprocessing.Process(target=compare_lists, args=(list1, list2, start_index, end_index, result_queue))
        p.start()
        processes.append(p)
    
    # 等待所有子进程完成
    for p in processes:
        p.join()
    
    # 从队列中获取每个进程返回的结果
    min_diff = float('inf')
    min_index = -1
    while not result_queue.empty():
        index, diff = result_queue.get()
        if diff < min_diff:
            min_diff = diff
            min_index = index
    
    return min_index

# 测试示例
list1 = [1, 3, 5, 7, 9, 11, 13, 15, 17, 19]
list2 = [2, 4, 6, 8, 10, 12, 14, 16, 18, 20]

min_index = compare_long_lists(list1, list2)
print("最小差值的索引为:", min_index)
print("list1中对应的元素为:", list1[min_index])
print("list2中对应的元素为:", list2[min_index])

此解决方法使用了多进程并行处理来加快对长列表的比较。首先，它获取CPU核心数并将长列表划分为多个块，每个块由一个子进程处理。每个子进程计算其分配的块中的最小差值和索引，并将结果放入进程间通信的队列中。最后，主进程从队列中获取每个子进程的结果，并找到最小差值和索引。

请注意，多进程并行处理适用于长列表，其中比较操作比较耗时。对于较小的列表，单线程处理可能更加高效。