LeetCode 190. Reverse Bits

原题链接在这里：https://leetcode.com/problems/reverse-bits/

题目：

Reverse bits of a given 32 bits unsigned integer.

Example 1:

Input: 00000010100101000001111010011100
Output: 00111001011110000010100101000000
Explanation: The input binary string 00000010100101000001111010011100 represents the unsigned integer 43261596, so return 964176192 which its binary representation is 00111001011110000010100101000000.

Example 2:

Input: 11111111111111111111111111111101
Output: 10111111111111111111111111111111
Explanation: The input binary string 11111111111111111111111111111101 represents the unsigned integer 4294967293, so return 3221225471 which its binary representation is 10111111111111111111111111111111.

Note:

Note that in some languages such as Java, there is no unsigned integer type. In this case, both input and output will be given as signed integer type and should not affect your implementation, as the internal binary representation of the integer is the same whether it is signed or unsigned.
In Java, the compiler represents the signed integers using 2's complement notation. Therefore, in Example 2 above the input represents the signed integer -3 and the output represents the signed integer -1073741825.

题解：

新学了一种API, Integer.reverse()可以直接返回binary的reverse. 参考链接：http://www.tutorialspoint.com/java/lang/integer_reverse.htm

Method 2 res左移一位后加上n的最右一位digit.

Time Complexity: O(1). 一共左移了32位.

Space O(1).

AC Java:

 1 public class Solution {
 2     // you need treat n as an unsigned value
 3     public int reverseBits(int n) {
 4         //Method 1
 5         //return Integer.reverse(n);
 6         
 7         //Method 2
 8         int res = 0;
 9         for(int i = 0; i<32; i++){
10             res <<= 1;
11             res += ((n>>i)&1);
12         }
13         return res;
14     }
15 }

Follow up 要求 this function is called many times.

把int 拆成 4个byte 分别reverse 反着加回去. 同时维护HashMap, key是byte, value是reverse 这个byte后的结果int.

若function 被call了很多次，那么大部分的byte都已经有了对应的reverse value存在HashMap中了。

AC Java:

 1 public class Solution {
 2     HashMap<Byte, Integer> hm = new HashMap<Byte, Integer>();
 3     
 4     public int reverseBits(int n) {
 5         byte [] bytes = new byte[4];
 6         
 7         //convert int into bytes
 8         for(int i = 0 ; i<4; i++){
 9             bytes[i] = (byte)((n>>>8*i) & 0xFF);
10         }
11         
12         int res = 0;
13         for(int i = 0; i<4; i++){
14             res <<= 8;
15             res += reverseByte(bytes[i]);
16         }
17         
18         return res;
19     }
20     
21     private int reverseByte(Byte b){
22         if(hm.containsKey(b)){
23             return hm.get(b);
24         }
25         
26         int val = 0;
27         for(int i = 0; i<8; i++){
28             val <<= 1;
29             val += ((b>>>i)&1);
30         }
31         hm.put(b, val);
32         return val;
33     }
34 }