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.
思路
每次左移一位,进行&运算。
代码
/*------------------------------------------------------
* 日期:2014-04-12
* 作者:SJF0115
* 题目: 191.Number of 1 Bits
* 网址:https://leetcode.com/problems/number-of-1-bits/
* 结果:AC
* 来源:LeetCode
* 博客:
--------------------------------------------------------*/
#include <iostream>
#include <vector>
using namespace std;
class Solution {
public:
int hammingWeight(uint32_t n) {
if(n == 0){
return 0;
}//if
if(n == 1){
return 1;
}//if
int count = 0;
for(int i = 0;i < 32;++i){
if((n&(1<<i)) > 0){
++count;
}//if
}//for
return count;
}
};
int main(){
Solution solution;
uint32_t num = 1;
cout<<""<<solution.hammingWeight(num)<<endl;
return 0;
}
思路二
每执行一次x = x&(x-1),会将x用二进制表示时最右边的一个1变为0,因为x-1将会将该位(x用二进制表示时最右边的一个1)变为0。
代码二
/*------------------------------------------------------
* 日期:2014-04-12
* 作者:SJF0115
* 题目: 191.Number of 1 Bits
* 网址:https://leetcode.com/problems/number-of-1-bits/
* 结果:AC
* 来源:LeetCode
* 博客:
--------------------------------------------------------*/
#include <iostream>
#include <vector>
using namespace std;
class Solution {
public:
int hammingWeight(int n) {
int count = 0;
while(n != 0){
n = n & (n-1);
count++;
}//while
return count;
}
};
int main(){
Solution solution;
uint32_t num = 11;
cout<<""<<solution.hammingWeight(num)<<endl;
return 0;
}
时间: 2024-11-17 19:45:07