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 as00111001011110000010100101000000).

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

Related problem: Reverse Integer

官方解答：

The Java solution is straightforward, just bitwise operation:

public int reverseBits(int n) {    int result = 0;    for (int i = 0; i < 32; i++) {        result += n & 1;        n >>>= 1;   // CATCH: must do unsigned shift        if (i < 31) // CATCH: for last digit, don't shift!            result <<= 1;    }    return result;}

How to optimize if this function is called multiple times? We can divide an int into 4 bytes, and reverse each byte then combine into an int. For each byte, we can use cache to improve performance.

// cacheprivate final Map<Byte, Integer> cache = new HashMap<Byte, Integer>();public int reverseBits(int n) {    byte[] bytes = new byte[4];    for (int i = 0; i < 4; i++) // convert int into 4 bytes        bytes[i] = (byte)((n >>> 8*i) & 0xFF);    int result = 0;    for (int i = 0; i < 4; i++) {        result += reverseByte(bytes[i]); // reverse per byte        if (i < 3)            result <<= 8;    }    return result;}private int reverseByte(byte b) {    Integer value = cache.get(b); // first look up from cache    if (value != null)        return value;    value = 0;    // reverse by bit    for (int i = 0; i < 8; i++) {        value += ((b >>> i) & 1);        if (i < 7)            value <<= 1;    }    cache.put(b, value);    return value;}