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.
思路:对[m,n]范围内所有的数取与,实际上最后结果就是m和n从最高前面有多少位同为1,然后把后面位全置零即可。
比如m=4;n=7
100&101&110=100
class Solution {
public:
int rangeBitwiseAnd(int m, int n) {
int i = 0;
while(m!=n){
m=m>>1;
n=n>>1;
++i;
}
return m<<i;
}
};