LeetCode题解：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).

题意：将一个整数的二进制表示逆转，得到逆转后的数

思路：转换不难，但题目要求优化该函数在多次调用时的性能表现，那就要做缓存了，如果只缓存某个数字，意义不大；但int是一个4字节数，我们可以缓存其字节

代码：

private 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;
}