Reverse Bits
Total Accepted: 54417 Total Submissions: 186739 Difficulty: Easy
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?
思路
public class Solution {
// you need treat n as an unsigned value
public int reverseBits(int n) {
int result = 0;
for (int i = 0; i < 32; i++) {
int a = (n >> i) & 1;
result = (result << 1) | a;
}
return result;
}
}
Follow up
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.
used the cache to improve performance
以4位为单位执行反转,将0x0至0xF的反转结果预存在一个长度为16的数组中,反转时直接查询即可。
// 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;
}