相似单词为 只差一位字母的单词，练习Map容器
package chapter4;
import java.util.*;
import java.util.Map.Entry;
/*
* 说明：找到一个单词的所有相似单词 例如： wine 和 dine wind 和wing 只有一个字母不同
*/
public class TreeMapTest {
/*
* 判断2个单词是否指差一个字母
*/
public static boolean oneCharOff(String s1, String s2) {
if (s1.length() != s2.length())
return false;
int diff = 0;
for (int i = 0; i < s1.length() - 1; i++) {
if (s1.charAt(i) != s2.charAt(i))
diff++;
if (diff > 1)
return false;
}
return diff == 1;
}
/*
* 打印方法
*/
public static void print(Map<String, List<String>> map) {
for (Map.Entry<String, List<String>> entry : map.entrySet()) {
List<String> words = entry.getValue();
System.out.print(entry.getKey() + ":");
for (int i = 0; i < words.size(); i++) {
System.out.print(words.get(i) + " ");
}
System.out.println();
}
}
/**
* 方法名：computeWords 说明：方法1
*/
public static Map<String, List<String>> computeWords1(List<String> words) {
Map<String, List<String>> map = new TreeMap<String, List<String>>();
String[] word = new String[words.size()];
words.toArray(word);
for (int i = 0; i < word.length; i++) {
for (int j = i + 1; j < word.length; j++) {
if (oneCharOff(word[i], word[j])) {
update(map, word[i], word[j]);// 互为相似单词
update(map, word[j], word[i]);
}
}
}
return map;
}
/**
* 方法名：update 说明：更新
*/
private static <KeyType> void update(Map<KeyType, List<String>> map,
KeyType key, String s) {
List<String> words = map.get(key);
if (words == null) {
words = new ArrayList<String>();
map.put(key, words);
}
words.add(s);
}
/**
* 方法名：groupByLength 说明：先将给的单词按照长度分组
*/
private static Map<Integer, List<String>> groupByLength(List<String> words) {
Map<Integer, List<String>> map = new TreeMap<Integer, List<String>>();
for (String s : words)
update(map, s.length(), s);
return map;
}
/**
* 方法名：computeWord 说明：方法2
*/
public static Map<String, List<String>> computeWords2(List<String> words) {
Map<String, List<String>> map = new TreeMap<String, List<String>>();
for (Entry<Integer, List<String>> entry : groupByLength(words)
.entrySet()) {
String[] word = new String[entry.getValue().size()];
entry.getValue().toArray(word);
for (int i = 0; i < word.length; i++) {
for (int j = i + 1; j < word.length; j++) {
if (oneCharOff(word[i], word[j])) {
update(map, word[i], word[j]);// 互为相似单词
update(map, word[j], word[i]);
}
}
}
}
return map;
}
/**
* 方法名：computeWords3
* 说明：该方法效率最高。首先也是按照长度分组，分组完了之后，对每组分别做以下操作：
* 1：从头到尾分别去掉每个单词的一位字母。将剩下的作为一个键，该单词作为值 放到新建的
* map<String,List<String>>reToWord里
* 2：遍历reToWord，找到size>=2的，(因为只有>=2的 才代表含有相似的)例如wine和wane 当去掉第2位
* 时，首先wine会进List,wane匹配到了wne也会进去，所以是2个
*/
public static Map<String, List<String>> computeWords3(List<String> words) {
Map<String, List<String>> adjWords = new TreeMap<String, List<String>>();
Map<Integer, List<String>> wordsByLength = new TreeMap<Integer, List<String>>();
for (String w : words)
update(wordsByLength, w.length(), w);
for (Map.Entry<Integer, List<String>> entry : wordsByLength.entrySet()) {
List<String> groupsWords = entry.getValue();
int groupNum = entry.getKey();
for (int i = 0; i < groupNum; i++) {
Map<String, List<String>> repToWord = new TreeMap<String, List<String>>();
for (String str : groupsWords) {
String rep = str.substring(0, i) + str.substring(i + 1);
update(repToWord, rep, str);
}
for (List<String> wordClique : repToWord.values())
if (wordClique.size() >= 2)
for (String s1 : wordClique)
for (String s2 : wordClique)
if (s1 != s2)
update(adjWords, s1, s2);
}
}
return adjWords;
}
public static void main(String[] args) {
// TODO Auto-generated method stub
List<String> list = new ArrayList<String>();
list.add("wane");
list.add("wine");
list.add("aine");
list.add("dine");
list.add("anew");
list.add("kine");
list.add("qine");
list.add("eine");
list.add("rine");
print(computeWords1(list));
System.out.println();
print(computeWords2(list));
System.out.println();
print(computeWords3(list));
}
}
原文地址：http://blog.csdn.net/xiuweikang/article/details/40626493
