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).

Follow up:

If this function is called many times, how would you optimize it?

Solution 1:

// you need treat n as an unsigned value
     public int reverseBits(int n) {
         int result=0;
         for(int i=0; i<32; i++){
             result = (result<<1)|(n&1);
             n = n >> 1;
         }
         return result;
     }

Optimized for repeat usage, solution 2:

// optimization idea: cache reverse numbers for all int numbers, but space complexity if O(2^32)
    // idea is to make reverse int into 4 reverse byte, in this case only cache O(2^8)
    public int reverseBits(int n) {
        byte[] temp = new byte[4];
        for(int i=0; i<4; i++, n=n>>>8){
            temp[i] = (byte)n;
        }
        int result=0;
        for(int i=0; i<4; i++){
            result = (result<<8) | ( reverseBytes(temp[i]) & 0xff );
        }
        return result;
    }
    public byte reverseBytes(byte n) {
        byte oldn=n;
        Byte result = cache.get(n);
        if(result!=null) return result;
        result=0;
        for(byte i=0; i<8; i++){
            result = (byte)((result<<1)|(n&1));
            n = (byte)(n>>1);
        }
        cache.put(oldn,result);
        return result;
    }