# [Leetcode] Bitwise and of Numbers Range

Source: Internet
Author: User
Tags bitwise

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, Inclus Ive.

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

Credits:
Special thanks to @amrsaqr for adding this problem and creating all test cases.

Has you met this question in a real interview? Idea: Find the law can, through analysis we found: 1, when m = N, directly return m or N. 2, when the length of the binary number representing m and n is unequal, the number of occurrences is bound to be 100 because the lower one is like a high one. 0, the length is a bit longer than the low number, this number and the length of the number is smaller than its "with", the result must be 0, 0 "and" any number is 0. That is, when the length of the binary number that represents m,n is inconsistent, the "and" operation is performed from M to N, the result is 0, and 3, when the length of the two is the same, the loop and operation can be done. Just note: When N is Int_max, special handling is required.
`classsolution{ Public:  intRangebitwiseand (intMintN) {intRET =m; intA = m, B =N; intLena =1, LenB =1; if(M = =N)returnm;  while(A >>=1) Lena++;  while(b >>=1) LenB++; if(Lena = =LenB) {       for(inti = m +1; I <= N; i++) {ret&=i; if(i = =Int_max) Break; }    }    Elseret=0; returnret; }};`

[Leetcode] Bitwise and of Numbers Range

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