1. 2的幂
231. 2 的幂 - 力扣（LeetCode）
class Solution {
public:
bool isPowerOfTwo(int n) {
return n>0&&n==(n&(-n));
}
};
2. 3的幂
326. 3 的幂 - 力扣（LeetCode）
class Solution {
public:
bool isPowerOfThree(int n) {
return n>0&&1162261467%n==0;
}
};
3. 数字范围按位与
201. 数字范围按位与 - 力扣（LeetCode）
class Solution {
public:
int rangeBitwiseAnd(int left, int right) {
while(left<right)
{
right-=right&(-right);
}
return right;
}
};
4. 颠倒二进制位
190. 颠倒二进制位 - 力扣（LeetCode）
class Solution {
public:
uint32_t reverseBits(uint32_t n) {
n=(n&0xaaaaaaaa)>>1|(n&0x55555555)<<1;
n=(n&0xcccccccc)>>2|(n&0x33333333)<<2;
n=(n&0xf0f0f0f0)>>4|(n&0x0f0f0f0f)<<4;
n=(n&0xff00ff00)>>8|(n&0x00ff00ff)<<8;
n=n>>16|n<<16;
return n;
}
};
5. 汉明距离
461. 汉明距离 - 力扣（LeetCode）
class Solution {
public:
int hammingDistance(int x, int y) {
int n=x^y;
n = (n & 0x55555555) + ((n >> 1) & 0x55555555);
n = (n & 0x33333333) + ((n >> 2) & 0x33333333);
n = (n & 0x0f0f0f0f) + ((n >> 4) & 0x0f0f0f0f);
n = (n & 0x00ff00ff) + ((n >> 8) & 0x00ff00ff);
n = (n & 0x0000ffff) + ((n >> 16) & 0x0000ffff);
return n;
}
};
6. 设计位集
2166. 设计位集 - 力扣（LeetCode）
class Bitset {
private:
std::vector<int> set;
const int size;
int zeros;
int ones;
bool reverse;
public:
Bitset(int n) : size(n), zeros(n), ones(0), reverse(false) {
set.resize((n + 31) / 32, 0);
}
void fix(int i) {
int index = i / 32;
int bit = i % 32;
if (!reverse) {
if ((set[index] & (1 << bit)) == 0) {
zeros--;
ones++;
set[index] |= (1 << bit);
}
} else {
if ((set[index] & (1 << bit)) != 0) {
zeros--;
ones++;
set[index] ^= (1 << bit);
}
}
}
void unfix(int i) {
int index = i / 32;
int bit = i % 32;
if (!reverse) {
if ((set[index] & (1 << bit)) != 0) {
ones--;
zeros++;
set[index] ^= (1 << bit);
}
} else {
if ((set[index] & (1 << bit)) == 0) {
ones--;
zeros++;
set[index] |= (1 << bit);
}
}
}
void flip() {
reverse = !reverse;
std::swap(zeros, ones);
}
bool all() const {
return ones == size;
}
bool one() const {
return ones > 0;
}
int count() const {
return ones;
}
std::string toString() const {
std::string result;
for (int k = 0, i = 0; i < size; k++) {
int number = set[k];
for (int j = 0; j < 32 && i < size; j++, i++) {
int status = (number >> j) & 1;
status ^= reverse ? 1 : 0;
result += std::to_string(status);
}
}
return result;
}
};
7. 两数相除
29. 两数相除 - 力扣（LeetCode）
class Solution {
public:
static const int MIN =INT_MIN;
int divide(int a, int b) {
if(a==MIN&&b==MIN)
{
return 1;
}
if(a!=MIN&&b!=MIN)
{
return div(a,b);
}
if(b==MIN)
{
return 0;
}
if(b==neg(1))
{
return INT_MAX;
}
a=add(a,b<0?neg(b):b);
int ans=div(a,b);
int offset=b<0?1:-1;
return add(ans,offset);
}
int div(int a, int b) {
int x = a < 0 ? neg(a) : a;
int y = b < 0 ? neg(b) : b;
int ans = 0;
for (int i = 30; i >= 0; i = minus(i, 1)) {
if ((x >> i) >= y) {
ans |= (1 << i);
x = minus(x, y << i);
}
}
return a < 0 ^ b < 0 ? neg(ans) : ans;
}
int add(int a,int b)
{
int ans=a;
while(b!=0)
{
ans=a^b;
b=(a&b)<<1;
a=ans;
}
return ans;
}
int minus(int a,int b)
{
return add(a,neg(b));
}
int neg(int n)
{
return add(~n,1);
}
int multiply(int a, int b) {
int ans = 0;
while (b != 0) {
if ((b & 1) != 0) {
ans = add(ans, a);
}
a <<= 1;
b >>= 1;
}
return ans;
}
};