Leetcode solution 190: Reverse Bits

标签：NPU   one   ora   where   ber   war   ota   htm   www   

Problem Statement 

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

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 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.

Problem link

Video Tutorial

You can find the detailed video tutorial here

• Youtube
• B站

Thought Process

Use this problem to bring awareness of bit manipulation in interviews, it‘s rare but sometimes do get asked by some interviewers.

We can directly use Java‘s Integer.reverse() method, not suitable for interview purpose.

Simple simulation, reading bits from the right most and adding them up while shifting left.

Regarding the follow up, caching is an obvious solution. There are two ways to cache

• Cache all the integers,  2^32 = 4294967296, total memory needed is 2^32*4B = 16GB roughly, easily handled by one server 
• Cache each byte, divide the 32 bits into 4 bytes. Memory footprint is only 2^8*1B = 256B, significantly smaller than 16GB

If you are curious on how Java SDK implemented this method, it uses the algorithm in Hackers Delight book section 7-1

 1195       /**
 1196        * Returns the value obtained by reversing the order of the bits in the
 1197        * two‘s complement binary representation of the specified {@code int}
 1198        * value.
 1199        *
 1200        * @return the value obtained by reversing order of the bits in the
 1201        *     specified {@code int} value.
 1202        * @since 1.5
 1203        */
 1204       public static int reverse(int i) {
 1205           // HD, Figure 7-1
 1206           i = (i & 0x55555555) << 1 | (i >>> 1) & 0x55555555;
 1207           i = (i & 0x33333333) << 2 | (i >>> 2) & 0x33333333;
 1208           i = (i & 0x0f0f0f0f) << 4 | (i >>> 4) & 0x0f0f0f0f;
 1209           i = (i << 24) | ((i & 0xff00) << 8) |
 1210               ((i >>> 8) & 0xff00) | (i >>> 24);
 1211           return i;
 1212       }

Solutions 

Simple simulation with follow up question

 1 public int reverseBits(int n) {
 2     int res = 0;
 3 
 4     for (int i = 0; i < 32; i++) {
 5         int t = n & 1;
 6         n = n >> 1;
 7         res = res << 1;
 8         res = res | t;
 9 
10     }
11     return res;
12     // return Integer.reverse(n);
13     }
14 
15 public int reverseBitsWithCache(int n) {
16     byte[] bytes = new byte[4];
17     Map<Byte, Integer> lookup = new HashMap<>();
18 
19     int reversedResult = 0;
20 
21     for (int i = 0; i < 4; i++) {
22         bytes[i] = (byte)((n >> (8 * i)) & 0xFF);
23         reversedResult = reversedResult << 8;
24         reversedResult = reversedResult | this.reverseByte(bytes[i], lookup);
25     }
26 
27     return reversedResult;
28 }
29 
30 public int reverseByte(byte a, Map<Byte, Integer> lookup) {
31     // TODO: validate input
32     if (lookup.containsKey(a)) {
33         return lookup.get(a);
34     }
35 
36     int reversedByte = 0;
37     for (int i = 0; i < 8; i++) {
38         int t = (a >> i) & 1;
39         reversedByte = reversedByte << 1;
40         reversedByte = reversedByte | t;
41     }
42 
43     lookup.put(a, reversedByte);
44 
45     return reversedByte;
46 }

Time Complexity: O(N), where N is number of bits in the input integer

Space Complexity: O(1), no extra space needed 

References

• Leetcode discussion solution

Leetcode solution 190: Reverse Bits

标签：NPU   one   ora   where   ber   war   ota   htm   www   

原文地址：https://www.cnblogs.com/baozitraining/p/12233711.html