你有一个整数s的集合,最初包含了从1到n的所有数字.不幸的是,由于某种错误,其中一个数字被重复到集合中的另一个数字,导致一个数字的重复和另一个数字的丢失.你得到一个整数数组数,代表该集合的数据状态.找到出现两次的数字和缺失的数字,并以数组的形式返回它们.在python中写出这个问题的解决方案.
Here is a possible solution to the problem in Python using a hash set:

```
def findErrorNums(nums):
    n = len(nums)
    seen = set()
    duplicated = missing = None
    
    for num in nums:
        if num in seen:
            duplicated = num
        else:
            seen.add(num)
    
    for i in range(1, n+1):
        if i not in seen:
            missing = i
            
    return [duplicated, missing]
```

The idea is to traverse the list and use a hash set to keep track of the numbers that have been seen so far. If we find a number that is already in the hash set, it means that it is duplicated. Otherwise, we add it to the hash set.

After that, we traverse the original list of integers from 1 to n, and for each number that is missing in the hash set, we assign it to the missing variable.

Finally, we return a list with the duplicated and missing numbers.

Note that this solution has a time complexity of O(n), where n is the length of the input list.