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

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

思路:序列是按1递增的,所以必定先影响低位,按位与为1的情况必定发生在高位。两个数向右移,当两数相等,剩余的1便是按位与得到的1。

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

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