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Write a function that takes an unsigned integer and returns the number of ’1‘ bits it has (also known as the Hamming weight).
For example, the 32-bit integer ’11‘ has binary representation 00000000000000000000000000001011, so the function should return 3.
class Solution {
public:
int hammingWeight(uint32_t n) {
int count = 0;
while (n)
{
++ count;
n = (n - 1) & n;
}
return count;
}
};比如容易想到的解决方案是下面这个,每次往右移动一位,判断最后一位是不是1,但这样会有点慢,32位就要移动32次。上面的方法巧妙的使用了(n-1) & n,这个有什么用?我们来看下具体的例子:
假设n= 1111000111000 那 n-1 = 1111000110111, (n-1) & n = 1111000110000,刚好把最后一个1给干掉了。也就是说, (n-1)&n 刚好会从最后一位开始,每次会干掉一个1.这样速度就比下面的快了。有几个1,执行几次。
class Solution {
public:
int hammingWeight(uint32_t n) {
int count = 0;
while(n)
{
if(n & 1)
count ++;
n = n >> 1;
}
return count;
}
};http://www.waitingfy.com/archives/1642
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原文地址:http://blog.csdn.net/fox64194167/article/details/44279511