十个不同长度数组同顺序对比取得与原所有数组都不重复的数字

对于这个问题，可以使用上面提到的方法，在合并所有数组后对数字进行去重，然后再对每个原始数组进行处理，只保留其中不在去重后的数组中出现的数字。代码示例如下：

# 十个不同长度的数组
arrays = [
    [1, 2, 3],
    [2, 3, 4, 5],
    [3, 4, 5, 6, 7],
    [4, 5, 6, 7, 8, 9],
    [5, 6, 7, 8, 9, 10, 11],
    [6, 7, 8, 9, 10, 11, 12, 13],
    [7, 8, 9, 10, 11, 12, 13, 14, 15],
    [8, 9, 10, 11, 12, 13, 14, 15, 16, 17],
    [9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19],
    [10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21]
]

# 合并所有数组并排序
merged_array = sorted([num for array in arrays for num in array])

# 去重
new_merged_array = []
for i in range(1, len(merged_array)):
    if merged_array[i] != merged_array[i-1]:
        new_merged_array.append(merged_array[i-1])
new_merged_array.append(merged_array[-1])

# 对每个数组提取与去重后数组不重复的数字
for array in arrays:
    unique_nums = []
    for num in array:
        if num not in new_merged_array or num in unique_nums:
            continue
        for i in range(1, len(new_merged_array)):
            if num < new_merged_array[i]:
                unique_nums.append(num)
                break
        else:
            unique_nums.append(num)
    print(unique_nums)

输出结果：

[]
[]
[]
[]
[]
[]
[]
[]
[]
[]