leetcode--Number of 1 Bits

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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.

解题思路：转字符或者借助容器，或者利用%

java:

package leedcode;public class numberOf1bit {    public static int hammingWeight(int n) {         int num = 0;            String[] s = Integer.toBinaryString(n).split("");//关键部分            for (int i = 0; i < s.length; i++){                if (s[i].equals("1"))                    num++;            }                       return num;    }    public static void main(String[] args) {        // TODO Auto-generated method stub        System.out.println(hammingWeight(00000000000000000000000000000010));            }}

java版本2：

package leedcode;public class numberOf1bit {    public static int hammingWeight(int n) {         int num = 0;            while(n!=0){            n&=n-1;//按位与            num++;            }               return num;    }    public static void main(String[] args) {        // TODO Auto-generated method stub        System.out.println(hammingWeight(00000000000000000000000000000010));            }}

java版本3：

public static int hammingWeight(int n) {            return Integer.bitCount(n);//本质还是位运算    }

c++版本1：

class Solution {public:    int hammingWeight(uint32_t n) {        bitset<32> num(n);        return num.count();    }};

c++版本2

class Solution {public:    int hammingWeight(uint32_t n) {        int num=0;        while(n!=0){        if(n%2==1)        num++;        n/=2}        return num;    }};

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