Description:
Given a positive integer, check whether it has alternating bits: namely, if two adjacent bits will always have different values.
Example 1:
Input: n = 5
Output: true
Explanation: The binary representation of 5 is: 101
Example 2:
Input: n = 7
Output: false
Explanation: The binary representation of 7 is: 111.
Example 3:
Input: n = 11
Output: false
Explanation: The binary representation of 11 is: 1011.
Example 4:
Input: n = 10
Output: true
Explanation: The binary representation of 10 is: 1010.
Example 5:
Input: n = 3
Output: false
Constraints:
1 <= n <= 231 - 1
C code solution:
bool hasAlternatingBits(int n){
int prev = n&1;
n = n >> 1;
while(n){
if(prev == (n&1)){
return false;
}
prev = n & 1;
n = n >> 1;
}
return true;
}
Description:
Reverse bits of a given 32 bits unsigned integer.
Example 1:
Input: n = 00000010100101000001111010011100
Output: 964176192 (00111001011110000010100101000000)
Explanation: The input binary string 00000010100101000001111010011100 represents the unsigned integer 43261596, so return 964176192 which its binary representation is 00111001011110000010100101000000.
Example 2:
Input: n = 11111111111111111111111111111101
Output: 3221225471 (10111111111111111111111111111111)
Explanation: The input binary string 11111111111111111111111111111101 represents the unsigned integer 4294967293, so return 3221225471 which its binary representation is 10111111111111111111111111111111.
Constraints:
- The input must be a binary string of length
32
C code solution:
uint32_t reverseBits(uint32_t n) {
int hash[32] = {0};
int i = 0;
// get the status of each digits into hash
while(n){
if(n&1 == 1)
hash[i] = 1;
i++;
n = n >> 1;
}
//reform the new number inversely
uint32_t res = 0;
for(i=31; i>=0; i--){
res += pow(2, 31-i) * hash[i];
}
return res;
}