letecode [190] - Reverse Bits

 Reverse bits of a given 32 bits unsigned integer.

 Example 1:

Input: 00000010100101000001111010011100
Output: 00111001011110000010100101000000
Explanation: The input binary string 00000010100101000001111010011100 represents the unsigned integer 43261596, so return 964176192 which its binary representation is 00111001011110000010100101000000.

Example 2:

Input: 11111111111111111111111111111101
Output: 10111111111111111111111111111111
Explanation: The input binary string 11111111111111111111111111111101 represents the unsigned integer 4294967293, so return 3221225471 which its binary representation is 10101111110010110010011101101001.

 

Note:

• Note that in some languages such as Java, there is no unsigned integer type. In this case, both input and output will be given as signed integer type and should not affect your implementation, as the internal binary representation of the integer is the same whether it is signed or unsigned.
• In Java, the compiler represents the signed integers using 2's complement notation. Therefore, in Example 2 above the input represents the signed integer -3 and the output represents the signed integer -1073741825.

Subject to the effect :

 　　Seeking a 32-bit unsigned binary integer inverse position.

Understanding:

 　　Method a: the indirect method

　　　　From the tail began to traverse the binary integer n. 1 n & n may calculate the last bit is 1 or 0; Update n = n >> 1.

　　　　The results are stored Retrograde with string. Each read in from the end of a string that is stored, when n = 0, if less than the length of the string 32 is filled with zeros.

　　　　Convert string to uint32_t type can be.

　　Method II: Direct Method

　　　　Trailing binary integer numbers is the first number after the reverse position, it is the inverse of its set value * pow (2, index), traversing from the tail n, the inverse value calculated directly opposed.

Code C ++:

method one:

class Solution {
public:
    uint32_t reverseBits(uint32_t n) {
        string str = "";
        int len;
        while(n!=0){
            str += to_string(n&1);
            n = n>>1;
        }
        len = str.length();
        while(len<32){
            str += '0';
            len++;
        }
        n = strtoul(str.c_str(),NULL,2);
        return n;
    }
};

Method Two:

class Solution {
public:
    uint32_t reverseBits(uint32_t n) {
        uint32_t sum=0;
        int index = 31;
        while(n!=0){
            if(n&1==1){
                sum += pow(2,index);    
            }
            index--;
            n = n>>1;
        }
        return sum;
    }
};

operation result:

 　　method one:

　　　　When execution: 4 ms, beat the 96.81% of all users to submit in C ++

　　　　Memory consumption: 8.3 MB, beat the 5.12% of all users to submit in C ++

　　Method Two:

　　　　When execution: 4 ms, beat the 96.81% of all users to submit in C ++

　　　　Memory consumption: 8.4 MB, beat the 5.12% of all users to submit in C ++