leetcode-190&191 Reverse Bits & Number of 1 Bits

一、190 Reverse Bits

Reverse bits of a given 32 bits unsigned integer.

For example, given input 43261596 (represented in binary as 00000010100101000001111010011100), return 964176192 (represented in binary as 00111001011110000010100101000000).

代码：

class Solution {
public:
uint32_t reverseBits(uint32_t n) {
stack<int> s;
int k=0;
while( n != 0)
{
s.push(n%2);//存储二进制码
n /= 2;
k++;//计算位数
}
uint32_t sum=0;
for(int i=k; i > 0; i--)
{
sum = sum + s.top()*pow(2 , 32-i);//逆顺序求值
s.pop();
}
return sum;
}
};

二、191 Number of 1 Bits.

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.

计算非负整数的二进制中1的个数。

时间8ms。

代码：

class Solution {
public:
int hammingWeight(uint32_t n) {
int k=0;
while(n != 0)
{
if( n % 2)
k++;
n /= 2;
}
return k;
}
};