{面试题10: 二进制中1的个数}

From 剑指Offer 何海涛 著

#include <iostream>
using namespace std;

int numberOf1(int n) {
    unsigned int newN = static_cast<unsigned int>(n); // 强制将 n 转换为一个无符号数, 避免负数左移, 产生死循环
    int count = 0;
    while(newN) {
        if(newN&1) {
            count++;
        }
        newN >>= 1;
    }
    return count;
}

int NumberOf1(int n) {
    int count = 0;
    while(n) {
        count++;
        n &= n-1; // 将 n 最右侧的 1 消耗掉
    }
    return count;
}

测试集:

void test(int n, int expected) {
    std::cout << std::boolalpha << (numberOf1(n) == expected) << std::endl;
    std::cout << std::boolalpha << (NumberOf1(n) == expected) << std::endl;
}
int main(int argc, char* argv[]) {
    test(0, 0);
    test(1, 1);
    test(10, 2);
    test(0x7fffffff, 31);
    test(0xffffffff, 32);
    test(0x80000000, 1);

    return 0;
}

总结:

1. 判断一个数 n 是否是 2 的幂?

(n != 0) && ((n & (n-1)) == 0)

2. 求不小于 n 的 2 的幂?

if(n>=1) {
    --n;
    n |= n >> 1;
    n |= n >> 2;
    n |= n >> 4;
    n |= n >> 8;
    n |= n >> 16;
    ++n;
}