[LeetCode] 191. Number of 1 Bits 二进制数1的个数

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.

写一个函数操作无符号整数，返回整数的二进制数1的个数。

解法：位操作Bit Manipulation

使用n&(n-1)的方法

假使 n =0x110101

n       n-1     n&(n-1)

step1: 110101 110100 110100

step2: 110100 110011 110000

step3: 110000 101111 100000

step4: 100000 011111 000000

发现有几个1，就循环几次n&(n-1)得到0。

时间复杂度：O(M),M是n中1的个数

Java:

public class Solution {
    public int NumberOf1(int n) {
        int count = 0;
        while (n != 0) {
            n &= (n - 1);
            count ++;
        }
        return count;
    }
}　　

Python:

class Solution:
    # @param n, an integer
    # @return an integer
    def hammingWeight(self, n):
        result = 0
        while n:
            n &= n - 1
            result += 1
        return result

Python:

class Solution(object):
    def hammingWeight(self, n):
        """
        :type n: int
        :rtype: int
        """
        res=0
        while n:
            res += (n&1)
            n >>= 1
        return res　　

Python:

class Solution(object):
    def hammingWeight(self, n):
        """
        :type n: int
        :rtype: int
        """
        b = bin(n)
        return b.count('1')

C++:

class Solution {
public:
    int hammingWeight(uint32_t n) {
        int res = 0;
        for (int i = 0; i < 32; ++i) {
            res += (n & 1);
            n = n >> 1;
        }
        return res;
    }
}; 

C++:

class Solution {
public:
    int hammingWeight(uint32_t n) {
        int count = 0;
        for (; n; n &= n - 1) {
            ++count;
        }
        return count;
    }
};

　

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[LeetCode] 190. Reverse Bits 翻转二进制位

　

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