位运算

// 位操作
// Page77
// 二进制中1的个数
/************
1.n & 1结果为1，则说明最右边的数为1；统计完 n右移一位，考虑下一位是否为1；
2.常规解法，每次循环把与n做与运算的1左移一位，不需要去改变原数n;
3.(n-1)&n的结果是把n的最右边的1变为0
===========
1循环次数是n中1的个数
2循环次数是整数二进制的位数,32为整数需要循环32次
3循环次数等于整数中1的个数
*************/
#include <iostream>
using namespace std;

void Solution1(int n){
    int cnt = 0;
    int cycle = 0;
    while (n){
        if (n & 1)
            cnt++;
        n = n >> 1; // 缺点：当n为一个负数 0x80000000，右移最高位补1，最终会变成0xFFFFFFFF陷入死循环
        cycle++;

    }
    //return cnt;
    cout << "Number of 1:" << cnt << endl;
    cout << "Number of cycle:" << cycle << endl;
}

int Solution2(int n){
    int cnt = 0;
    unsigned int flag = 1;
    int cycle = 0;
    while (flag){
        if (n & flag)
            cnt++;
        flag = flag << 1;
        cycle++;
    }
    //return cnt;
    cout << "Number of 1:" << cnt << endl;
    cout << "Number of cycle:" << cycle << endl;
}

int Solution3(int n){
    int cnt = 0;
    int cycle = 0;
    while (n){
        cnt++;
        n = (n - 1) & n;
        cycle++;
    }
    cout << "Number of 1:" << cnt << endl;
    cout << "Number of cycle:" << cycle << endl;
}


int main(){
    unsigned int x = 0x7FFFFFFF;
    //int y = 0x80000000;
    int y = 0xFFFFFFFF;
    int z = 0;
    Solution1(z);
    Solution2(z);
    Solution3(z);
    return 0;

}

leetcode中一道关于倒置位的题reverse bits

/********************************
Reverse bits of a given 32 bits unsigned integer.

For example, given input 43261596 (represented in binary as 00000010100101000001111010011100), 
   return 964176192 (represented in binary as 00111001011110000010100101000000).
**********************************/
class Solution {
public:
    uint32_t reverseBits(uint32_t n) {
        uint32_t result = 0;
        for (int i = 0; i < 32; i++){
            result = (result << 1) + ((n >> i) & 1); 
        }
        return result;
    }
};