Analysis of leetcode notes in Bitwise AND of Numbers Range

Analysis of leetcode notes in Bitwise AND of Numbers Range

I. Description

Given a range[m, n]Where0 <= m <= n <= 2147483647, Return the bitwise AND of all numbers in this range, inclusive.

For example, given the range[5, 7], You shoshould return4.

Ii. Question Analysis

Given a range[m, n], Where0 <= m <= n <= 2147483647Returns the bitwise AND of All integers in the range, including the boundary.mAndn. For example, if the specified range is[5, 7], Should return4.

By looking at the description of the question, you can know that this question involves in-place operations. Therefore, you need to perform an analysis:

For the range [5, 7] given by the question, it can be written:

5: 1016: 110  -> 100 -> 47: 111

Here are two examples:

The value range is [6, 8]:

6: 01107: 0111  -> 0000 -> 08: 1000

The value range is [8, 10]:

8:  10009:  1001  -> 1000 -> 410: 1010

Now we can summarize the rule, for the range[m, n], IfnThe highest effective bit ratiomIf the maximum valid bit of is high, there will inevitably be two numbers in the range, which are the reverse codes of the other Party:

6: 01107: 0111  -> 0000 -> 08: 1000

Therefore, only whenmAndnWhen the maximum valid bits are the same, that is, m and n are shifted right at the same time, andm == n，m == 1When the bitwise AND result are not 0, the result ismAndnIn the left-side header of the public, the following provides simple code:

Iii. Sample Code

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

Iv. Summary

This is a mathematical problem that can be quickly solved using bitwise operations, mainly to find out the law.