201. Bitwise AND of Numbers Range

Given a range [m, n] where 0 <= m <= n <= 2147483647, return the bitwise AND of all numbers in this range, inclusive.

For example, given the range [5, 7], you should return 4.

题意：返回m到n之间所有数的与

思路：低位有不同，则低位与的结果为0，去除低位的不同，直到m n相等的时候，说明此后到位都是相同的。返回这个结果即可。

class Solution {public:int rangeBitwiseAnd(int m, int n) {int p = 0;while (m != n){m >>= 1;n >>= 1;p++;}return m << p;}};