# 201. 数字范围按位与

<https://leetcode-cn.com/problems/bitwise-and-of-numbers-range/>

## 解法一：

思路是找出m和n**相同**和**不同**的部分。相同部分与之后结果不变，不同部分与结果为0。每次n，m左移一位，直到n==m，用move来统计移动的位数（不同的位数）。最后用m\*move还原（即给m补0）。

```python
class Solution:
    def rangeBitwiseAnd(self, m: int, n: int) -> int:
        #边界，不加也行，不过慢
        if m == 0:
            return 0
        move = 1
        while n != m:
            n >>= 1
            m >>= 1
            move <<= 1
        return m * move
```


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