算法： Reverse Bits 反转位收藏

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？
A：这其实才是这道题的精髓，考察的大规模数据时算法最基本的优化方法。其实道理很简单，反复要用到的东西记下来就行了，所以我们用Map记录之前反转过的数字和结果。更好的优化方法是将其按照Byte分成4段存储，节省空间。参见这个帖子。

 

// cache
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; }
转自：https://segmentfault.com/a/1190000003483740