当存在2个以上的对时，解决合并和排序对问题

def merge_sorted_arrays(arrays):
    merged = []
    # 初始化指针
    pointers = [0] * len(arrays)
    
    while any(pointers[i] < len(arrays[i]) for i in range(len(arrays))):
        min_val = float('inf')
        min_idx = -1
        for i, pointer in enumerate(pointers):
            if pointer < len(arrays[i]) and arrays[i][pointer] < min_val:
                min_val = arrays[i][pointer]
                min_idx = i
        merged.append(min_val)
        pointers[min_idx] += 1
    
    return merged

# 示例
arrays = [[1, 4, 5], [1, 3, 4], [2, 6]]
print(merge_sorted_arrays(arrays))  # 输出: [1, 1, 2, 3, 4, 4, 5, 6]

def merge_sorted_arrays(arrays):
    if len(arrays) == 1:
        return arrays[0]
    
    mid = len(arrays) // 2
    left = merge_sorted_arrays(arrays[:mid])
    right = merge_sorted_arrays(arrays[mid:])
    
    return merge(left, right)

def merge(left, right):
    merged = []
    i = j = 0
    
    while i < len(left) and j < len(right):
        if left[i] < right[j]:
            merged.append(left[i])
            i += 1
        else:
            merged.append(right[j])
            j += 1
    
    merged.extend(left[i:])
    merged.extend(right[j:])
    
    return merged

# 示例
arrays = [[1, 4, 5], [1, 3, 4], [2, 6]]
print(merge_sorted_arrays(arrays))  # 输出: [1, 1, 2, 3, 4, 4, 5, 6]