英文:
LeetCode: count number of 1 bits
问题
我正在尝试理解为什么下面的代码在这个问题中不起作用。
我们将位（0和1）作为参数传递。如果我不使用位运算，而是首先将整数转换为charArray，然后迭代它以计算'1'的数量，最后返回它，为什么它不起作用？
public class Solution {
// 你需要将n视为无符号值
public int hammingWeight(int n) {
int count=0;
for(char c:String.valueOf(n).toCharArray())
{
if('1'==c)
++count;
}
return count;
}
}
英文:
I am trying to understand why below code won't work for this problem.
We are passing bits (0's & 1's) as an argument. If I do not use Bit operations and rather I first convert the integer to charArray and iterate over it to count the no of '1' and then return it, why it does not work?
public class Solution {
// you need to treat n as an unsigned value
public int hammingWeight(int n) {
int count=0;
for(char c:String.valueOf(n).toCharArray())
{
if('1'==c)
++count;
}
return count;
}
}
答案1
得分: 1
你正在一个十进制字符串中计数1，你可以通过Integer.toBinaryString()轻松将其转换为二进制字符串。
public int hammingWeight(int n) {
int count = 0;
for (char c : Integer.toBinaryString(n).toCharArray()) {
if ('1' == c)
++count;
}
return count;
}
英文:
You are counting 1 in a decimal string, you can eaily convert it to a binary string by Integer.toBinaryString()
public int hammingWeight(int n) {
int count=0;
for(char c:Integer.toBinaryString(n).toCharArray())
{
if('1'==c)
++count;
}
return count;
}
答案2
得分: 0
鉴于这是Java，我很惊讶没有人提到过java.lang.Integer的JDK方法Integer.bitCount(i)。
英文:
Given this is Java, I am surprised nobody mentioned the JDK method Integer.bitCount(i) of java.lang.Integer.
答案3
得分: -1
我们不需要将n转换为字符串，这将通过得很好：
class Solution {
public static int hammingWeight(int n) {
int ones = 0;
while (n != 0) {
ones = ones + (n & 1);
n = n >>> 1;
}
return ones;
}
}
英文:
We don't have to convert the n to string, this would pass just fine:
class Solution {
public static int hammingWeight(int n) {
int ones = 0;
while (n != 0) {
ones = ones + (n & 1);
n = n>>>1;
}
return ones;
}
}
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