1. Word Count
题目描述：
Using map reduce to count word frequency
样例：
chunk1: "Google Bye GoodBye Hadoop code"
chunk2: "lintcode code Bye"
Get MapReduce result:
Bye: 2
GoodBye: 1
Google: 1
Hadoop: 1
code: 2
lintcode: 1
代码实现：
/**
* Definition of OutputCollector:
* class OutputCollector {
* public void collect(K key, V value);
* // Adds a key/value pair to the output buffer
* }
*/
public class WordCount {
public static class Map {
public void map(String key, String value, OutputCollector output) {
// Write your code here
// Output the results into output buffer.
// Ps. output.collect(String key, int value);
StringTokenizer tokenizer = new StringTokenizer(value);
while (tokenizer.hasMoreTokens()) {
String word = tokenizer.nextToken();
output.collect(word, 1);
}
}
}
public static class Reduce {
public void reduce(String key, Iterator values,
OutputCollector output) {
// Write your code here
// Output the results into output buffer.
// Ps. output.collect(String key, int value);
int sum = 0;
while (values.hasNext()) {
sum += values.next();
}
output.collect(key, sum);
}
}
}
2. 反向索引
题目描述：
创建给定文档的反向索引
样例：
给出一个包括id与内容的文档list(我们提供了document类)
[
{
"id": 1,
"content": "This is the content of document 1 it is very short"
},
{
"id": 2,
"content": "This is the content of document 2 it is very long bilabial bilabial heheh hahaha ..."
},
]
返回一个反向索引(hashmap的key是单词, value是文档的id).
{
"This": [1, 2],
"is": [1, 2],
...
}
代码实现：
/**
* Definition of Document:
* class Document {
* public int id;
* public String content;
* }
*/
public class Solution {
/**
* @param docs a list of documents
* @return an inverted index
*/
public Map> invertedIndex(List docs) {
// Write your code here
Map> results = new HashMap>();
for (Document doc : docs) {
int id = doc.id;
StringBuffer temp = new StringBuffer("");
String content = doc.content;
int n = content.length();
for (int i = 0; i < n; ++i) {
if (content.charAt(i) == ' ') {
insert(results, temp.toString(), id);
temp = new StringBuffer("");
} else
temp.append(content.charAt(i));
}
insert(results, temp.toString(), id);
}
return results;
}
public void insert(Map> rt, String tmp, int id) {
if (tmp.equals("") || tmp == null)
return;
if (!rt.containsKey(tmp))
rt.put(tmp, new ArrayList());
int n = rt.get(tmp).size();
if (n == 0 || rt.get(tmp).get(n - 1) != id)
rt.get(tmp).add(id);
}
}
3. 倒排索引
题目描述：
使用 map reduce 来实现一个倒排索引
样例：
下面是要被索引的文本：
T0:"it is what it is"
T1:"what is it"
T2:"it is a banana"
我们就能得到下面的反向文件索引：
"a": {2}
"banana": {2}
"is": {0, 1, 2}
"it": {0, 1, 2}
"what": {0, 1}
代码实现：
/**
* Definition of OutputCollector:
* class OutputCollector {
* public void collect(K key, V value);
* // Adds a key/value pair to the output buffer
* }
* Definition of Document:
* class Document {
* public int id;
* public String content;
* }
*/
public class InvertedIndex {
public static class Map {
public void map(String _, Document value,
OutputCollector output) {
// Write your code here
// Output the results into output buffer.
// Ps. output.collect(String key, int value);
int id = value.id;
StringTokenizer tokenizer = new StringTokenizer(value.content);
while (tokenizer.hasMoreTokens()) {
String word = tokenizer.nextToken();
output.collect(word, id);
}
}
}
public static class Reduce {
public void reduce(String key, Iterator values,
OutputCollector> output) {
// Write your code here
// Output the results into output buffer.
// Ps. output.collect(String key, List value);
List results = new ArrayList();
int previous = -1;
while (values.hasNext()) {
int now = values.next();
if(previous != now) {
results.add(now);
}
previous = now;
}
output.collect(key, results);
}
}
}
4. GFS客户端
题目描述：
为GFS(Google文件系统)实现一个简单的客户端，提供一下功能：
1.read(文件名)，通过文件名从GFS中读取文件。
2.write(文件名，内容)，通过文件名和内容写入GFS中。
现在有两种已经在基础类中实现的方法：
1.readChunk(文件名，块索引)，从GFS中读取一个块。
2.writeChunk(文件名，块索引，块数据)，向GFS中写入一个块。
为了简化这个问题，我们可以假设块大小为 chunkSize 位的(在真实的文件系统中，是64M)，GFS客户端的任务是将一个文件分为若干块(如果需要的话)并且保存在远端的GFS服务器上，chunkSize会在构造函数中给出，你需要的是实现读和写这两个private方法。
样例：
GFSClient(5)
read("a.txt")
>> null
write("a.txt", "World")
>> You don't need to return anything, but you need to call writeChunk("a.txt", 0, "World") to write a 5 bytes chunk to GFS.
read("a.txt")
>> "World"
write("b.txt", "111112222233")
>> You need to save "11111" at chink 0, "22222" at chunk 1, "33" at chunk 2.
write("b.txt", "aaaaabbbbb")
read("b.txt")
>> "aaaaabbbbb"
代码实现：
/* Definition of BaseGFSClient
* class BaseGFSClient {
* private Map chunk_list;
* public BaseGFSClient() {}
* public String readChunk(String filename, int chunkIndex) {
* // Read a chunk from GFS
* }
* public void writeChunk(String filename, int chunkIndex,
* String content) {
* // Write a chunk to GFS
* }
* }
*/
public class GFSClient extends BaseGFSClient {
public int chunkSize;
public Map chunkNum;
public GFSClient(int chunkSize) {
// initialize your data structure here
this.chunkSize = chunkSize;
this.chunkNum = new HashMap();
}
// @param filename a file name
// @return conetent of the file given from GFS
public String read(String filename) {
// Write your code here
if (!chunkNum.containsKey(filename))
return null;
StringBuffer content = new StringBuffer();
for (int i = 0; i < chunkNum.get(filename); ++i) {
String sub_content = readChunk(filename, i);
if (sub_content != null)
content.append(sub_content);
}
return content.toString();
}
// @param filename a file name
// @param content a string
// @return void
public void write(String filename, String content) {
// Write your code here
int length = content.length();
int num = (length - 1) / chunkSize + 1;
chunkNum.put(filename, num);
for (int i = 0; i < num; ++i) {
int start = i * chunkSize;
int end = i == num -1 ? length : (i + 1) * chunkSize;
String sub_content = content.substring(start, end);
writeChunk(filename, i, sub_content);
}
}
}
5. 最常使用的k个单词(Map Reduce)
题目描述
使用map reduce框架查找最常使用的k个单词.
mapper的key为文档的id, 值是文档的内容, 文档中的单词由空格分割.
对于reducer，应该输出最多为k个key-value对, 包括最常用的k个单词以及他们在当前reducer中的使用频率.评判系统会合并不同的reducer中的结果以得到 全局 最常使用的k个单词, 所以你不需要关注这一环节. k 在TopK类的构造器中给出.
样例
给出文档 A =
lintcode is the best online judge
I love lintcode
以及文档 B =
lintcode is an online judge for coding interview
you can test your code online at lintcode
最常用的2个单词以及他们的使用频率应该为:
lintcode, 4
online, 3
代码实现
class Pair {
String key;
int value;
Pair(String key, int value) {
this.key = key;
this.value = value;
}
}
public class TopKFrequentWords {
public static class Map {
public void map(String _, Document value,
OutputCollector output) {
// Write your code here
// Output the results into output buffer.
// Ps. output.collect(String key, int value);
int id = value.id;
String content = value.content;
String[] words = content.split(" ");
for (String word : words)
if (word.length() > 0) {
output.collect(word, 1);
}
}
}
public static class Reduce {
private PriorityQueue Q = null;
private int k;
private Comparator pairComparator = new Comparator() {
public int compare(Pair left, Pair right) {
if (left.value != right.value) {
return left.value - right.value;
}
return right.key.compareTo(left.key);
}
};
public void setup(int k) {
// initialize your data structure here
this.k = k;
Q = new PriorityQueue(k, pairComparator);
}
public void reduce(String key, Iterator values) {
// Write your code here
int sum = 0;
while (values.hasNext()) {
sum += values.next();
}
Pair pair = new Pair(key, sum);
if (Q.size() < k) {
Q.add(pair);
} else {
Pair peak = Q.peek();
if (pairComparator.compare(pair, peak) > 0) {
Q.poll();
Q.add(pair);
}
}
}
public void cleanup(OutputCollector output) {
// Output the top k pairs into output buffer.
// Ps. output.collect(String key, Integer value);
List pairs = new ArrayList();
while (!Q.isEmpty()) {
pairs.add(Q.poll());
}
// reverse result
int n = pairs.size();
for (int i = n - 1; i >= 0; --i) {
Pair pair = pairs.get(i);
output.collect(pair.key, pair.value);
}
}
}
}