1829. Maximum XOR for Each Query
## 思路
先對整個nums陣列做XOR
k = xor ^ mask (11111111)
for loop紀錄k並更新xor值
## Code
```python
class Solution:
def getMaximumXor(self, nums: List[int], maximumBit: int) -> List[int]:
n = len(nums)
xor = 0
for num in nums:
xor ^= num
res = []
mask = (1 << maximumBit) - 1
for i in range(n-1, -1, -1):
res.append(xor ^ mask)
xor ^= nums[i]
return res
```

大師