题目:
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.
在《剑指offer》上看到过这个题目,通过右移操作逐位和1进行与运算来统计题目中的无符号整型数字的32位二进制形式中包含的1的个数。
代码:
public class Solution {
// you need to treat n as an unsigned value
public int hammingWeight(int n) {
int count = 0;
int i;
for(i = 0; i < 32; i++){
if((n&1) == 1){
count++;
}
n = n >> 1;
}
return count;
}
}
【注】
在《剑指offer》中看到的题目并没有限定是非负数,因此如果像上述代码一样遇到输入是负数的情况会发生死循环。