- Leetcode 190. Reverse Bits
- Leetcode 191. Number of 1 Bits
- Leetcode 338. Counting Bits
- Leetcode 371. Sum of Two Integers
Bit Series II
Leetcode 190. 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) {
int res = 0;
for (int i=0; i<32; i++) {
res += n&1;
n>>=1;
if (i<31) res <<=1;
}
return res;
}
};
Leetcode 191. Number of 1 Bits
Write a function that takes an unsigned integer and returns the number of ’1’ bits it has (also known as the Hamming weight).
For example, the 32-bit integer ’11’ has binary representation 00000000000000000000000000001011, so the function should return 3.
class Solution {
public:
int hammingWeight(uint32_t n) {
int res = 0;
while (n) {
if (n%2==1) res++;
n /=2;
}
return res;
}
};
Leetcode 338. Counting Bits
Given a non negative integer number num. For every numbers i in the range 0 ≤ i ≤ num calculate the number of 1’s in their binary representation and return them as an array.
Example:
For num = 5 you should return [0,1,1,2,1,2].
class Solution {
public:
vector<int> countBits(int num) {
vector<int> res(num+1, 0);
for (int i=1; i<=num; i++)
res[i] = res[i>>1]+i%2;
return res;
}
};
Leetcode 371. Sum of Two Integers
Calculate the sum of two integers a and b, but you are not allowed to use the operator + and -.
Example:
Given a = 1 and b = 2, return 3.
class Solution {
public:
int getSum(int a, int b) {
int res = a^b;
int carray = (a&b)<<1;
if (carray!=0) return getSum(res, carray);
return res;
}
};