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.
Credits:
Special thanks to @amrsaqr for adding this problem and creating all test cases.
思路:我们先以[26,30]为例,有:
11010 11011 11100 11101 11110
可见最后求得的结果是范围内所有的数公共的二进制前缀,即最终求得是最长公共二进制前缀。我们可以每次将m和n右移一位,直至m,n相等,同时记录向右移动的次数i,此时m就是我们要求的公共二进制前缀右移i位得到的数,将m左移i位即可求得最终结果。
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class Solution { public: int rangeBitwiseAnd(int m, int n) { int offset=0; while(m!=n){ m>>=1; n>>=1; offset++; } return m<<offset; } };