[Leetcode 190] Reverse Bits

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;
}