Leetcode-190. inverting binary bits

An operation

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Topic details

Invert given 32 The binary bit of an unsigned integer .
Tips ：
Please note that , In some languages （ Such as Java） in , There is no unsigned integer type . under these circumstances , Both input and output will be specified as signed integer types , And should not affect your implementation , Because whether integers are signed or unsigned , Its internal binary representation is the same .
stay Java in , The compiler uses binary complement notation to represent signed integers . therefore , stay Example 2 in , The input represents a signed integer -3, The output represents a signed integer -1073741825.

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Example 1：

 Input ：n = 00000010100101000001111010011100
 Output ：964176192 (00111001011110000010100101000000)
 explain ： Binary string of input  00000010100101000001111010011100  Represents an unsigned integer  43261596,
      Therefore return  964176192, Its binary representation is  00111001011110000010100101000000.

Example 2:

 Input ：n = 11111111111111111111111111111101
 Output ：3221225471 (10111111111111111111111111111111)
 explain ： Binary string of input  11111111111111111111111111111101  Represents an unsigned integer  4294967293,
      Therefore return  3221225471  Its binary representation is  10111111111111111111111111111111 .

Ideas :
Use arithmetic shift left and right , You can flip , Every time res Move left and fill on the right 0, And then take n The last one is filled in res
Then make n Move right , In order to remove one , for instance :

My code ：

class Solution 
{
    
public:
    uint32_t reverseBits(uint32_t n) 
    {
    
        uint32_t res = 0;
        for (int i = 0; i < 32; ++i)
        {
    
            res <<= 1; // Answer shift left 
            res += n & 1; // take n last place 
            n >>= 1;   //n Move right 
        } 
        return res;
    }
};

You can also use divide and conquer , Divide the binary string into two parts , Exchange the left and right parts , Then use the same method to deal with the left and right parts respectively until all parts are exchanged , That is, the flip is completed

class Solution 
{
    
public:
    uint32_t reverseBits(uint32_t n) 
    {
    
        n = (n >> 16) | (n << 16);
        n = ((n & 0xff00ff00) >> 8) | ((n & 0x00ff00ff) << 8);
        n = ((n & 0xf0f0f0f0) >> 4) | ((n & 0x0f0f0f0f) << 4);
        n = ((n & 0xcccccccc) >> 2) | ((n & 0x33333333) << 2);
        n = ((n & 0xaaaaaaaa) >> 1) | ((n & 0x55555555) << 1);
        return n;
    }
};

Common skills of bit operation