java-并查集（Disjoint-set）-将多个集合合并成没有交集的集合
import java.util.ArrayList;import java.util.Arrays;import java.util.HashMap;import java.util.HashSet;import java.util.Iterator;import java.util.List;import java.util.Map;import java.util.Set;public class DisjointSet {/**题目：给定一个字符串的集合，格式如：{aaa bbb ccc}， {bbb ddd}，{eee fff}，{ggg}，{ddd hhh}要求将其中交集不为空的集合合并，要求合并完成后的集合之间无交集，例如上例应输出{aaa bbb ccc ddd hhh}，{eee fff}， {ggg}。（1）请描述你解决这个问题的思路；（2）请给出主要的处理流程，算法，以及算法的复杂度（3）请描述可能的改进。解答：1. 假定每个集合编号为0，1，2，3...2. 创建一个hash_map，key为字符串，value为一个链表，链表节点为字符串所在集合的编号。遍历所有的集合，将字符串和对应的集合编号插入到hash_map中去。3. 创建一个长度等于集合个数的int数组，表示集合间的合并关系。例如，下标为5的元素值为3，表示将下标为5的集合合并到下标为3的集合中去。开始时将所有值都初始化为-1，表示集合间没有互相合并。在集合合并的过程中，我们将所有的字符串都合并到编号较小的集合中去。 遍历第二步中生成的hash_map，对于每个value中的链表，首先找到最小的集合编号（有些集合已经被合并过，需要顺着合并关系数组找到合并后的集合编号），然后将链表中所有编号的集合都合并到编号最小的集合中（通过更改合并关系数组）。4.现在合并关系数组中值为-1的集合即为最终的集合，它的元素来源于所有直接或间接指向它的集合。0: {aaa bbb ccc}1: {bbb ddd}2: {eee fff}3: {ggg}4: {ddd hhh}生成的hash_map，和处理完每个值后的合并关系数组分别为aaa: 0 [-1, -1, -1, -1, -1]bbb: 0, 1 [-1, 0, -1, -1, -1]ccc: 0 [-1, 0, -1, -1, -1]ddd: 1, 4 [-1, 0, -1, -1, 0]eee: 2 [-1, 0, -1, -1, 0]fff: 2 [-1, 0, -1, -1, 0]ggg: 3 [-1, 0, -1, -1, 0]hhh: 4 [-1, 0, -1, -1, 0]所以合并完后有三个集合，第0，1，4个集合合并到了一起，第2，3个集合没有进行合并。Use "Disjoin-set".But I use "HashSet" and "HashMap" of Java API.Does "Disjoin-set" have its own data structure?see also [url]http://www.csie.ntnu.edu.tw/~u91029/DisjointSets.html[/url] */private final int SIZE=7;private int[] father;//the root in disjion set.private static List<Set<String>> resultList=new ArrayList<Set<String>>();public static void main(String[] args) {String[] str0={"aaa","bbb","ccc",};String[] str1={"bbb","ddd",};String[] str2={"eee","fff",};String[] str3={"ggg",};String[] str4={"ddd","hhh",};String[] str5={"xx","yy",};String[] str6={"zz","yy",};String[][] strs={str0,str1,str2,str3,str4,str5,str6};//change String[][] to List<Set>for(String[] str:strs){//when I write--"Arraylist list=Arrays.asList(strArray)","addAll()" is unsupported for such a arraylist.Set<String> set=new HashSet<String>();set.addAll(Arrays.asList(str));resultList.add(set);}DisjointSet disjointSet=new DisjointSet();disjointSet.disjoin(strs);}public void disjoin(String[][] strings){if(strings==null||strings.length<2)return;initial();Map<String,List<Integer>> map=storeInHashMap(strings);union(map);}//in the beginning,each element is in its own "group".public void initial(){father=new int[SIZE];for(int i=0;i<SIZE;i++){father[i]=i;}}/*Map<k,v> * key:String * value:List<Integer>-in which sets the string shows up. */public Map<String,List<Integer>> storeInHashMap(String[][] strings){Map<String,List<Integer>> map=new HashMap<String,List<Integer>>();for(int i=0;i<SIZE;i++){for(String each:strings[i]){if(!map.containsKey(each)){List<Integer> list=new ArrayList<Integer>();list.add(i);map.put(each, list);}else{map.get(each).add(i);}}}//traverse the hashmapIterator<Map.Entry<String, List<Integer>>> it=map.entrySet().iterator();while(it.hasNext()){Map.Entry<String, List<Integer>> entry=it.next();String key=entry.getKey();List<Integer> value=entry.getValue();System.out.println(key+":"+value);}return map;}public void union(Map<String,List<Integer>> map){Iterator<Map.Entry<String, List<Integer>>> it=map.entrySet().iterator();while(it.hasNext()){Map.Entry<String, List<Integer>> entry=it.next();List<Integer> value=entry.getValue();unionHelp(value);//the arrays whose indexes are in the same list should be merged to one set.}System.out.println("the father array is "+Arrays.toString(father));//merge two setsfor(int i=0;i<SIZE;i++){if(i!=father[i]){Set<String> dest=resultList.get(father[i]);Set<String> source=resultList.get(i);dest.addAll(source);}}//clear a set which has been added.for(int i=0;i<SIZE;i++){if(i!=father[i]){resultList.get(i).clear();}}System.out.println("after merge:"+resultList);}public void unionHelp(List<Integer> list){int minFather=getFather(list.get(0));//list[0] is the smaller.for(int i=0,size=list.size();i<size;i++){father[list.get(i)]=minFather;}}//general union in disjoin set.But we overload it in this case.public void unionHelp(int x,int y){if(father[x]!=father[y]){int fx=getFather(x);int fy=getFather(y);//merge two arrays to the array that has a smaller index.if(fx<fy){father[y]=fx;}else{father[x]=fy;}}}public int getFather(int x){while(x!=father[x]){x=father[x];}return x;}}