计算长整数中的置位位数 [英] Counting number of set bits in a long number

问题描述

此问题已在此处得到了回答.

This question has been answered here.

我的查询是，按照方法1起作用，但是它的变化，即方法2无效，而是给出了预期输出值的两倍.我不知道为什么.

My query is, following approach-1 works, however the variation of it, that is approach-2 does not, rather it gives double the value of expected output. I can not find out why.

方法1

public class Solution {
    public int numSetBits(long a) {
        int count = 0;
        long temp = 0;
        for(int i = 0 ; i < 64 ; i++) { // 64-bit for long data-type
            temp = 1;
            temp = temp << i;
            temp = a & temp;
            if((temp > 0))
                count++;
        }
      return count;
    }
}

方法2

public class Solution {
    public int numSetBits(long a) {
        int count=0;
        for(int i=0; i<64; i++) {
            if((a & (1 << i))>0) count++;
        }
      return count;
    }
}

推荐答案

第二种方法失败，因为1 << i的结果是int，而不是long.因此，位掩码回绕了，因此a的低32位被扫描了两次，而a的高32位被忽略了.

The second approach fails because the result of 1 << i is an int, not a long. So the bit mask wraps around, and so the lower 32 bits of a get scanned twice, while the higher 32 bits of a are left uncounted.

因此，当i达到值32时，(1 << i)不会是2 ³² ，而是2 ⁰ (即1)，与当i为0时.类似地，当i为33时，(1 << i)不会是2 ³³ ，而是2 ¹ ，...等等.

So, when i reaches the value 32, (1 << i) will not be 2³², but 2⁰ (i.e. 1), which is the same as when i was 0. Similarly when i is 33, (1 << i) will not be 2³³, but 2¹, ...etc.

通过将常数1设为long来纠正此问题:

Correct this by making the constant 1 a long:

(1L << i)

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