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?

移位法

复杂度

时间 O(1) 空间 O(1)

思路

最简单的做法，原数不断右移取出最低位，赋给新数的最低位后新数再不断左移。

代码

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

分段相或法

复杂度

时间 O(1) 空间 O(1)

思路

Java标准的Integer.reverse()源码。

代码

public class Solution {    // you need treat n as an unsigned value    public int reverseBits(int i) {        i = (i & 0x55555555) << 1 | (i >>> 1) & 0x55555555;        i = (i & 0x33333333) << 2 | (i >>> 2) & 0x33333333;        i = (i & 0x0f0f0f0f) << 4 | (i >>> 4) & 0x0f0f0f0f;        i = (i << 24) | ((i & 0xff00) << 8) | ((i >>> 8) & 0xff00) | (i >>> 24);        return i;    }}

后续 Follow Up

Q：如果该方法被大量调用，或者用于处理超大数据（Bulk data）时有什么优化方法？

// cacheprivate final Map
    
      cache = new HashMap
     
      ();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;}