求大神会java的hashmap的问题:如何删掉Map中重复的值？

Question

如：map.put("1","abc");map.put("2","abc");map.put("3","kjs");map.put("4","abc");map("5","kjs");
运行后2,4,5对应的map都会从list中删掉。
由于不太懂java，求详细代码，会加分

Answer 1

import java.util.HashSet;
import java.util.Iterator;
import java.util.Map;
import java.util.Map.Entry;
import java.util.Set;

import org.apache.commons.collections4.map.LinkedMap;

public class Test030 {

    /**
     * @param args
     */
    public static void main(String[] args) {
        // TODO Auto-generated method stub
        Map<String, String> m = new LinkedMap<String, String>();
        m.put("1", "abc");
        m.put("2", "abc");
        m.put("3", "kjs");
        m.put("4", "abc");
        m.put("5", "kjs");
        System.out.println("before: " + m);
        removeDuplicate(m);
        System.out.println("after: " + m);
    }

    private static void removeDuplicate(Map<String, String> m) {
        Set<String> values = new HashSet<String>();
        for (Iterator<Entry<String, String>> it = m.entrySet().iterator(); it
                .hasNext();) {
            Entry<String, String> e = it.next();
            if(values.contains(e.getValue())){
                it.remove();
            }else{
                values.add(e.getValue());
            }
        }
    }

}

 

写的时候没有注意, 如果jdk中没有linkedMap, 那么请引入commons的collections包

主要是HashMap不保证顺序

追问

非常感谢你的回答，不过我水平不够，你的这个代码我没用得起来。

Answer 2

package test;

import java.util.HashMap;
import java.util.Iterator;
import java.util.Map;
import java.util.TreeMap;

import org.eclipse.jdt.internal.compiler.ast.DoStatement;

public class A {
    public static void main(String[] args) {
        Map<String, String> map = new HashMap<String, String>();
        map.put("1","abc");
        map.put("2","abc");
        map.put("3","kjs");
        map.put("4","abc");
        map.put("5","kjs");
        Map<String, String> map4 = new HashMap<String, String>();
        map4 = doThing(map);
        for (String key : map4.keySet()){
            System.out.println(key+":"+map4.get(key));
        }

    }
    public static Map<String, String>  doThing(Map<String, String> map){
        Map<String, String> map2 = new HashMap<String, String>();
        Map<String, String> map3 = new HashMap<String, String>();
        //TreeMap:对map按key值排序
        TreeMap<String, String> treemap = new TreeMap<String, String>(map);
        Iterator<String> it = treemap.keySet().iterator();
        while (it.hasNext()) {
            String key = it.next();
            String value = treemap.get(key);
            if(map2.containsKey(value)){
                continue;
            }else{
                map2.put(value, value);
                map3.put(key, value);
            }
            
        }
        return map3;
    }
}

输出： 

3:kjs

1:abc

本回答被提问者采纳

Answer 3

HashMap 是一个散列表，它存储的内容是键值对(key-value)映射。
HashMap
继承于AbstractMap，实现了Map、Cloneable、java.io.Serializable接口。
HashMap
的实现不是同步的，这意味着它不是线程安全的。它的key、value都可以为null。此外，HashMap中的映射不是有序的。它存储的值是无序不可重复的！

Answer 4

给你个例子看，就知道了
package shuai.study.map;

import java.util.ArrayList;
import java.util.Collections;
import java.util.Comparator;
import java.util.HashMap;
import java.util.Iterator;
import java.util.List;
import java.util.Map;
import java.util.Map.Entry;
import java.util.Set;
import java.util.TreeMap;

/**
* @author shengshu
*
*/
public class UniqueMap {

// Remove repetition from Map, this is core part in this Class
public static Map<String, String> removeRepetitionFromMap(Map<String, String> map) {
Set<Entry<String, String>> set = map.entrySet();

List<Entry<String, String>> list = new ArrayList<Entry<String, String>>(set);

Collections.sort(list, new Comparator<Entry<String, String>>() {
@Override
public int compare(Entry<String, String> entry1, Entry<String, String> entry2) {
return Integer.valueOf(entry1.getValue().hashCode()) - Integer.valueOf(entry2.getValue().hashCode());
}
});

// list.size() is dynamic change
for (int index = 0; index < list.size(); index++) {
String key = list.get(index).getKey();
String value = list.get(index).getValue();

int next_index = index + 1;

if (next_index < list.size()) {
String next_key = list.get(next_index).getKey();
String next_value = list.get(next_index).getValue();

// Remove repetition record whose key is more bigger
if (value == next_value) {
if (key.hashCode() < next_key.hashCode()) {
map.remove(next_key);
list.remove(next_index);
} else {
map.remove(key);
list.remove(index);
}

// Due to having repetition in List, so index will be reduced
index--;
}
}
}

return map;
}

// Transfer Map to Sorted Map
public static Map<String, String> transferToSortedMap(Map<String, String> map) {
// Define comparator for TreeMap
Map<String, String> new_sort_map = new TreeMap<String, String>(new Comparator<String>() {
@Override
public int compare(String key1, String key2) {
return key1.hashCode() - key2.hashCode();
}
});

new_sort_map.putAll(map);

return new_sort_map;
}

public static void printMap(Map<String, String> map) {
Iterator<Entry<String, String>> iterator = map.entrySet().iterator();

while (iterator.hasNext()) {
Entry<String, String> entry = iterator.next();

String key = entry.getKey();
String value = entry.getValue();

System.out.println(key + " --> " + value);
}
}

public static void main(String[] args) {
Map<String, String> map = new HashMap<String, String>();
map.put("A", "1");
map.put("B", "2");
map.put("C", "2");
map.put("D", "3");
map.put("E", "3");

Map<String, String> new_map = UniqueMap.removeRepetitionFromMap(map);

// new_sort_map is what we want
Map<String, String> new_sort_map = UniqueMap.transferToSortedMap(new_map);

// Print new_sort_map
UniqueMap.printMap(new_sort_map);
}
}

Answer 5

把键值对调