Baozi Leetcode 190: Reverse Bits & 中年大叔程序员沮丧聊天疫情中的WFH

包子面试培训

发布于 2020-05-12 15:20:15

4800

发布于 2020-05-12 15:20:15

文章被收录于专栏：包子铺里聊IT

包子小道消息@05/09/2020

哈喽大家好！有段时间没有跟大家更新了，各位疫情期间在家工作是否顺心？欢迎收听包子聊天系列：中年程序员大叔沮丧聊天-疫情中WFH怎么感觉更累了？！

油管：https://youtu.be/hFIc9Z2uSAA
B站：https://www.bilibili.com/video/BV17a4y1i7Yu/

包子小道消息

还是各大公司雷人的消息，Lyft 982, Airbnb 1900，Uber也动手了3700的非Tech部门，接下来还会有tech部门的layoff。大家互相比较severance package也可以间接的反应各个公司的文化，都是资本家，都要逐利，看怎么样执行才会格调高一些
疫情着实影响了今年毕业找工作的同学们，老美也还好，不行回家住父母地下室，中国留学生在opt和visa的压力下就更加艰难了。大家可以去https://layoffs.fyi/tracker/ 里面也有哪些公司招人的信息。国际大气候和湾区局部小气候造成了目前的局势，地主家现在也没有余粮了，先拿个包裹，别考虑有多lowball，之后再说。

[photo credit: Danielle Davis]

Blogger:https://blog.baozitraining.org/2020/03/leetcode-solution-190-reverse-bits.html

Youtube: https://youtu.be/PYS5s26t26U

博客园: https://www.cnblogs.com/baozitraining/p/12233711.html

B站: https://www.bilibili.com/video/av85055139/

Problem Statement

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.

Follow up:

If this function is called many times, how would you optimize it?

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 }