June 9th training day - bit operation

Preface

The content of today's algorithm is ： An operation

One 、 Binary spacing

         Given a positive integer n, Find and return to n In the binary representation of adjacent 1 Between Longest distance . If There are no two adjacent 1, return 0.
If only 0 Put two 1 Separate （ May not exist 0 ）, We think these two 1 each other adjacent . Two 1 The distance between them is the absolute difference of their positions in the binary representation . for example ,“1001” Two of them 1 The distance to 3 .

One 、 Ideas :

        （1） Use a variable to indicate whether it exists in binary 1 And store the previous 1, Use another variable to store the real-time location （ Since it is a binary number, there is no need to worry about whether there will be Yes 1 The previous case is zero , It will end the cycle directly ）
        （2） Make a circular judgment , If it conforms to 1, Do it once. 1 And 1 Judging the length between And Assignment of existing variables , If not, the length will be shifted to the right and the real-time variable will be increased ;

Two 、 Source code

class Solution {
    
public:
    int binaryGap(int n) {
    
        int maxd=0;
        int pre=-1,now=0;
        while(n){
    
            if(n&1){
    
                if(pre!=-1){
    
                    maxd=max(now-pre,maxd);
                }
                pre=now;
            }
            n>>=1;
            ++now;
        }
        return maxd;
    }
};
// Hero's code 
// I wanted to  n To binary stored in the array for a double pointer judgment , Not very familiar hh

3、 ... and . Knowledge point

         An operation & ,>>

Today is a contestant …