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;
}