这道题的输入是一个无符号的整数
有两种办法
解法一、暴力解
把整数变成字符串,遍历字符串查找1的个数,注意,这里不能用String.valueOf
public class Solution {
// you need to treat n as an unsigned value
public int hammingWeight(int n) {
String str = Integer.toBinaryString(n);
int count = 0;
for(int i = 0; i < str.length();i++){
char a = str.charAt(i);
if(a == '1')
count++;
}
return count;
}
}
解法二、位运算
用31个0和1个1和每一位做与操作,如果该位是1则结果为1,否则为0
与操作的规则是两个都是1则为1,否则为0
public class Solution {
// you need to treat n as an unsigned value
public int hammingWeight(int n) {
int res = 0;
int mask = 1;
for(int i = 0; i < 32; i++){
if ((n & mask) != 0) {
res++;
}
mask<<=1;
}
return res;
}