編寫一個函式，輸入是一個無符號整數，返回其二進位制表示式中數字位數為 ‘1’ 的個數（也被稱為漢明重量）。

示例 1：

輸入：00000000000000000000000000001011
輸出：3
解釋：輸入的二進位制串 00000000000000000000000000001011 中，共有三位為 '1'。

示例 2：

輸入：00000000000000000000000010000000
輸出：1
解釋：輸入的二進位制串 00000000000000000000000010000000 中，共有一位為 '1'。

示例 3：

輸入：11111111111111111111111111111101
輸出：31
解釋：輸入的二進位制串 11111111111111111111111111111101 中，共有 31 位為 '1'。

提示：

• 請注意，在某些語言（如 Java）中，沒有無符號整數型別。在這種情況下，輸入和輸出都將被指定為有符號整數型別，並且不應影響您的實現，因為無論整數是有符號的還是無符號的，其內部的二進位制表示形式都是相同的。
• 在 Java 中，編譯器使用二進位制補碼記法來表示有符號整數。因此，在上面的 示例 3 中，輸入表示有符號整數 -3。

進階:
如果多次呼叫這個函式，你將如何優化你的演算法？

解題思路

類似於之前Leetcode 190：顛倒二進位制位（超詳細的解法！！！）問題，稍加修改即可。

class Solution(object):
    def hammingWeight(self, n):
        """
        :type n: int
        :rtype: int
        """
        res = 0
        for
 
 _ in range(32):
            res += n & 1
            n >>= 1
            
        return res

使用python語言的話，一個取巧的做法。

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

這個問題的高階玩法就是漢明重量（Hamming Weight）

//types and constants used in the functions below

typedef unsigned __int64 uint64;  //assume this gives 64-bits
const uint64 m1 = 0x5555555555555555; //binary: 0101...
const uint64 m2 = 0x3333333333333333; //binary: 00110011..
const uint64 m4 = 0x0f0f0f0f0f0f0f0f; //binary:  4 zeros,  4 ones ...
const uint64 m8 = 0x00ff00ff00ff00ff; //binary:  8 zeros,  8 ones ...
const uint64 m16 = 0x0000ffff0000ffff; //binary: 16 zeros, 16 ones ...
const uint64 m32 = 0x00000000ffffffff; //binary: 32 zeros, 32 ones ...
const uint64 hff = 0xffffffffffffffff; //binary: all ones
const uint64 h01 = 0x0101010101010101; //the sum of 256 to the power of 0,1,2,3...

//This is a naive implementation, shown for comparison,
//and to help in understanding the better functions.
//It uses 24 arithmetic operations (shift, add, and).
int popcount_1(uint64 x) {
    x = (x & m1 ) + ((x >>  1) & m1 ); //put count of each  2 bits into those  2 bits 
    x = (x & m2 ) + ((x >>  2) & m2 ); //put count of each  4 bits into those  4 bits 
    x = (x & m4 ) + ((x >>  4) & m4 ); //put count of each  8 bits into those  8 bits 
    x = (x & m8 ) + ((x >>  8) & m8 ); //put count of each 16 bits into those 16 bits 
    x = (x & m16) + ((x >> 16) & m16); //put count of each 32 bits into those 32 bits 
    x = (x & m32) + ((x >> 32) & m32); //put count of each 64 bits into those 64 bits 
    return x;
}

//This uses fewer arithmetic operations than any other known  
//implementation on machines with slow multiplication.
//It uses 17 arithmetic operations.
int popcount_2(uint64 x) {
    x -= (x >> 1) & m1;             //put count of each 2 bits into those 2 bits
    x = (x & m2) + ((x >> 2) & m2); //put count of each 4 bits into those 4 bits 
    x = (x + (x >> 4)) & m4;        //put count of each 8 bits into those 8 bits 
    x += x >>  8;  //put count of each 16 bits into their lowest 8 bits
    x += x >> 16;  //put count of each 32 bits into their lowest 8 bits
    x += x >> 32;  //put count of each 64 bits into their lowest 8 bits
    return x &0xff;
}

//This uses fewer arithmetic operations than any other known  
//implementation on machines with fast multiplication.
//It uses 12 arithmetic operations, one of which is a multiply.
int popcount_3(uint64 x) {
    x -= (x >> 1) & m1;             //put count of each 2 bits into those 2 bits
    x = (x & m2) + ((x >> 2) & m2); //put count of each 4 bits into those 4 bits 
    x = (x + (x >> 4)) & m4;        //put count of each 8 bits into those 8 bits 
    return (x * h01)>>56;  //returns left 8 bits of x + (x<<8) + (x<<16) + (x<<24) + ... 
}

使用cpp的話有兩個高階的玩法。

int hammingWeight(uint32_t n) 
{
    bitset<32> bin(n);
    return bin.count();
}

更高階的技巧

return __builtin_popcount(n);

不過這種做法僅對GCC編譯器有效。

reference:

https://en.wikipedia.org/wiki/Hamming_weight

我將該問題的其他語言版本新增到了我的GitHub Leetcode

如有問題，希望大家指出！！！