长位的Java位操作 - 计数设置和未设置位 [英] Java Bit Operation on Long - Counting Set and Unset bits

问题描述

我的号码很长。现在我想要的是（以伪代码给出），

I have a long number. Now what I want is following (given in pseudo code) ,

int cnt1 = 0 
int cnt2 = 0 

for each two bits of that long

       if the two bits == 11
            then cnt1++

       else
            cnt2++

    Print i and i+1 th bits are ... (example 00, 11 etc.) and cnt1 = ... and cnt2 = ...

(for example if number is three (representation "00 00 00 .... 11)" 
  it will give output cnt1 = 1 and cnt2 = 31)

有人可以帮我怎么做吗？

Can anybody help me how to do that ?

推荐答案

你需要什么do是在每次迭代时向右移动2位，并使用数字3（二进制11）进行按位和（&）操作：

What you need to do is keep shifting by 2 bits to the right at every iteration and do a bitwise and (&) operation with the number 3 (11 in binary):

long number;
int cnt1 = 0;
int cnt2 = 0;
long test = 3;
int counter = 0;    

while(counter < 64) { // we have 64 bits to inspect
    if((number & test) == 3) { // last 2 bits are 11
        cnt1++;
    } else { // last 2 bits are either 10, 01 or 00
        cnt2++;
    }          
    counter += 2;
    number = number >>> 2; // shift by 2 bits to the right
}

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