位运算
1 // 位操作 2 // Page77 3 // 二进制中1的个数 4 /************ 5 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； 6 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; 7 3.(n-1)&n的结果是把n的最右边的1变为0 8 =========== 9 1循环次数是n中1的个数 10 2循环次数是整数二进制的位数,32为整数需要循环32次 11 3循环次数等于整数中1的个数 12 *************/ 13 #include <iostream> 14 using namespace std; 15 16 void Solution1(int n){ 17 int cnt = 0; 18 int cycle = 0; 19 while (n){ 20 if (n & 1) 21 cnt++; 22 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 23 cycle++; 24 25 } 26 //return cnt; 27 cout << "Number of 1:" << cnt << endl; 28 cout << "Number of cycle:" << cycle << endl; 29 } 30 31 int Solution2(int n){ 32 int cnt = 0; 33 unsigned int flag = 1; 34 int cycle = 0; 35 while (flag){ 36 if (n & flag) 37 cnt++; 38 flag = flag << 1; 39 cycle++; 40 } 41 //return cnt; 42 cout << "Number of 1:" << cnt << endl; 43 cout << "Number of cycle:" << cycle << endl; 44 } 45 46 int Solution3(int n){ 47 int cnt = 0; 48 int cycle = 0; 49 while (n){ 50 cnt++; 51 n = (n - 1) & n; 52 cycle++; 53 } 54 cout << "Number of 1:" << cnt << endl; 55 cout << "Number of cycle:" << cycle << endl; 56 } 57 58 59 int main(){ 60 unsigned int x = 0x7FFFFFFF; 61 //int y = 0x80000000; 62 int y = 0xFFFFFFFF; 63 int z = 0; 64 Solution1(z); 65 Solution2(z); 66 Solution3(z); 67 return 0; 68 69 }
leetcode中一道关于倒置位的题reverse bits
1 /******************************** 2 Reverse bits of a given 32 bits unsigned integer. 3 4 For example, given input 43261596 (represented in binary as 00000010100101000001111010011100),
return 964176192 (represented in binary as 00111001011110000010100101000000). 5 **********************************/ 6 class Solution { 7 public: 8 uint32_t reverseBits(uint32_t n) { 9 uint32_t result = 0; 10 for (int i = 0; i < 32; i++){ 11 result = (result << 1) + ((n >> i) & 1); 12 } 13 return result; 14 } 15 };
