java的Hashmap中排序问题_百度知道
java的Hashmap中排序问题
相对通话次数排序，怎么进行啊，通话次序也定义成了StringHashmap不是只能存储String类型数据吗
提问者采纳
但是你存数值的时候，取出来的时候你用String的方式取出来，然后你转化成数值类型的就行了。用Integer.parseInt(res)在hashmap中存什么类型都可以
提问者评价
来自团队：
其他类似问题
为您推荐：
其他3条回答
不一定能第一个取出来Hashmap的key和value都可以存储Object类型的，并且Hashmap是不保证存储顺序的，不仅仅是String，也就是说第一个存进去的值
HashMap里面的对象要实现一个接口 implements Comparable&T& 这样就可以用ComparTo方法进行自定义排序了
谁给你说的Hashmap只能存String类型的？
hashmap的相关知识
等待您来回答
下载知道APP
随时随地咨询
出门在外也不愁java list&map& 排序_百度知道
java list&map& 排序
map22= list,Object&OMap&& maptemp=new HashMap&().get(&
Map& j++) {
Map& list=new ArrayList&
Map&lt, Object& map=new HashMap&lt, Object&Map&();O
int map2value=(Integer) list.get(j+1), Object&number&number&(); list1=new ArrayList&Map&lt.add(map);;;;O map2=new HashMap&
if (map1value&map2value) {
maptemp=map11;Object,Object&
Map&&gt.add(map3), Object&gt,Object&gt.size()-i-1;number&quot.get(j);(); map22;& i++) {
for (int j = 0;
list, 2););O();Object,Object&gt.get(j+1);O
map11=map22.put(&Ofor (int i = 0.size().put(&
map11= list, Object&&Object, Object&O
Map&); i &
map3;Object, Object&number&quot, Object&;Onumber&quot.get(list.add(map2);OObject.get(&
list1, Object&(); j &Map&lt, 1), 4); map3=new HashMap&lt.add(map1).get(j); map11, Object&
Map&lt.put(&number&O map1=new HashMap&lt.add(list.size()-i)).put(&quot, Object&gt, Object&O
int map1value=(Integer)
map2, 3);()public static void main(String[] args) {
哪里错了呢
提问者采纳
&nbsp,&&nbsp,&HashMap&&&&&&nbsp.add(map);&Map&&&&&&public&&ArrayList&&&&static&&&&nbsp.HashM&&nbsp.&Object&o2;&;&Object&&&&&&=&nbsp.add(map2);java.&&&=&number&&&&O&&nbsp.util.put(&new&nbsp.put(&&&&&&&&&&nbsp.Mimport&&(),&return&&nbsp.sort(&&nbsp.CO&nbsp.).put(&&nbsp,&&&nbsp,&&nbsp,&&&3);&&&compare(Map&number&&&&&&&&&map2;&map2&import&&args)&nbsp,&&S&&number&&&&&&&&&&&&;&&map1value&number&O&Object&Onumber&quot.ArrayL=&HashMap&public&nbsp,&&O&map3;&&&map&Map&&&Object&gt.add(map1);&nbsp,&&&import&o2)&nbsp.get(&&main(String[]&nbsp,&&&();public&&&&nbsp.&&map1&&();Map&list&&&&&C&O(Integer)&&&new&List&int&&{&nbsp.add(map3);&&&class&&&O&&&O&&O{&&&nbsp.HashMap&=&map1value&new&&&(Integer)&int&Object&;&O&&&import&nbsp,&$&O&&4);&Object&&&&}&&&&});Object&&&&&nbsp.put(&&import&&nbsp,&&{&&O&&&&nbsp.&nbsp,&Map&&2);int&);&&nbsp,&o1;&&&&&nbsp,&map1;{&&&&map2value&&&=&&()&&&new&&&nbsp,&&&&&nbsp,&Object&&&&&number&Object&&map2value.L&nbsp,&&&void&new&nbsp用我这个排序不是更简单么import&Object&&&();&Object&&&&&Map&Comparator&&1);=&Map&Object&O&nbsp.println(list);;&o1;&&Map&HashMap&&=&&&&&Object&&();&nbsp.get(&&&new&&nbsp,&&&Map&&map3&&-&&&nbsp
您回答的对，现在我想问下用冒泡排序的话，怎么才能写出来呢？谢谢你
不好意思，才看见，你这个不适合用冒泡排序做。实现起来非常麻烦，冒泡的，建议你理解的概念，就可以了。上面代码的30行~40行，用下面的代替&&&&&&&&System.out.println(list);&&&&&&&&for&(int&i&=&0;&i&&&list.size();&i++)&{&&&&&&&&&&&&Map&Object,&Object&&tmp0&=&list.get(i);&&&&&&&&&&&&int&number0&=&(Integer)&tmp0.get(&number&);&&&&&&&&&&&&Map&Object,&Object&&tmp&=&&&&&&&&&&&&&for&(int&j&=&i;&j&&&list.size();&j++)&{&&&&&&&&&&&&&&&&Map&Object,&Object&&tmp1&=&list.get(j);&&&&&&&&&&&&&&&&int&number1&=&(Integer)&tmp1.get(&number&);&&&&&&&&&&&&&&&&if&(number0&&&number1)&{&&&&&&&&&&&&&&&&&&&&tmp&=&tmp0;&&&&&&&&&&&&&&&&&&&&list.set(i,&list.get(j));&&&&&&&&&&&&&&&&&&&&list.set(j,&tmp);&&&&&&&&&&&&&&&&}&&&&&&&&&&&&}&&&&&&&&}&&&&&&&&System.out.println(list);&这个是用冒泡排序实现的，你对比下吧，哪个简单？你出现的问题是：对象间用等号赋值，那么就让这两个对象指向了同一个内存地址，也就是等同的，即，可以理解为对东一个文件创建两个快捷方式，但实际指向的还是同一个文件。利用list里面的set方法来赋值
提问者评价
太厉害了，不知道怎么给你加点分。非常感谢
来自团队：
其他类似问题
为您推荐：
java的相关知识
其他1条回答
带Tree的类多好啊，排序方便啊
等待您来回答
下载知道APP
随时随地咨询
出门在外也不愁&p&一堆键值均重复的键值对，去掉值重复的键值对，并记录值被重复的信息， 把值和信息封装成对象。与键再构造键值对。有时会报错。&img src=&/feaf64bd7baaf538e7d5b_b.jpg& data-rawwidth=&1852& data-rawheight=&972& class=&origin_image zh-lightbox-thumb& width=&1852& data-original=&/feaf64bd7baaf538e7d5b_r.jpg&&&br&&/p&&p&public static IdentityHashMap&String, JSONObject& removeRepetitionFromMap(IdentityHashMap&String, JSONObject& map) {&/p&&p&
System.setProperty(&java.util.Arrays.useLegacyMergeSort&, &true&);&/p&&p&
int i = 0 ;&/p&&p&
Set&Entry&String, JSONObject&& set = map.entrySet();&/p&&p&
List&Entry&String, JSONObject&& list = new ArrayList&Entry&String, JSONObject&&(set);&/p&&p&
Collections.sort(list, new Comparator&Entry&String, JSONObject&&() {&/p&&p&
public int compare(Entry&String, JSONObject& entry1, Entry&String, JSONObject& entry2) {&/p&&p&
return Integer.valueOf(entry1.getValue().hashCode()) - Integer.valueOf(entry2.getValue().hashCode());&/p&&p&
for (int index = 0; index & list.size(); index++) {&/p&&p&
String key = list.get(index).getKey();&/p&&p&
JSONObject value = list.get(index).getValue();&/p&&p&
int next_index = index + 1;&/p&&p&
if (next_index & list.size()) {&/p&&p&
String next_key = list.get(next_index).getKey();&/p&&p&
JSONObject next_value = list.get(next_index).getValue();&/p&&p&
if (value.toString().equals(next_value.toString())) {&/p&&p&
Dup = &/p&&p&
if (key.hashCode() & next_key.hashCode()) {&/p&&p&
map.remove(next_key);&/p&&p&
list.remove(next_index);
} else {&/p&&p&
map.remove(key);&/p&&p&
list.remove(index);
index--;&/p&&p&
duplicateRequestArray.add(value);&/p&&p&
} else{&/p&&p&
Dup =&/p&&p&
ArrayList&JSONObject& jsonList = new ArrayList&JSONObject&();&/p&&p&
Iterator&Entry&String, JSONObject&& iterator = map.entrySet().iterator();&/p&&p&
while (iterator.hasNext()) {&/p&&p&
Entry&String, JSONObject& entry = iterator.next();
JSONObject value = entry.getValue();&/p&&p&
jsonList.add(value);&/p&&p&
if(map.size()!=(removeDuplicate(jsonList).size())){
removeRepetitionFromMap(map);&/p&&p&
一堆键值均重复的键值对，去掉值重复的键值对，并记录值被重复的信息， 把值和信息封装成对象。与键再构造键值对。有时会报错。public static IdentityHashMap&String, JSONObject& removeRepetitionFromMap(IdentityHashMap&String, JSONObject& map) {
System.setProperty("java.util.Arrays.useLegacyMergeSort", "true");
int i = 0 ;
Set&Entry&String, JSONObject&& set = map.entrySet();
List&Entry&String, JSONObject&& list = new ArrayList&Entry&String, JSONObject&&(set);
Collections.sort(list, new Comparator&Entry&String, JSONObject&&() {
public int compare(Entry&String, JSONObject& entry1, Entry&String, JSONObject& entry2) {
你这个犯了初学者常见的错误，边遍历java容器list，map，set边删除的错误。正确的做法是遍历的时候记录下要删除的数据，之后再统一删除。参考下这个代码吧：
已有帐号？
无法登录？
社交帐号登录Visit the Apple Store
locations.
1-800-MY-APPLE
Copyright & 2011 Apple Inc. All rights reserved.Java: how to convert HashMap&String, Object& to array - Stack Overflow
to customize your list.
Stack Overflow is a community of 4.7 million programmers, just like you, helping each other.
J it only takes a minute:
Join the Stack Overflow community to:
Ask programming questions
Answer and help your peers
Get recognized for your expertise
I need to convert a HashMap&String, Object& could anyone show me how it's done?
86.6k34250402
14.4k153450
hashMap.keySet().toArray(); // returns an array of keys
hashMap.values().toArray(); // returns an array of values
Should be noted that the ordering of both arrays may not be the same,
See oxbow_lakes answer for a better approach for iteration when the pair key/values are needed.
18.7k3278109
If you want the keys and values, you can always do this via the entrySet:
hashMap.entrySet().toArray(); // returns a Map.Entry&K,V&[]
From each entry you can (of course) get both the key and value via the getKey and getValue methods
86.6k34250402
If you have HashMap&String, SomeObject& hashMap then
hashMap.keySet().toArray();
will return an Object[]. If instead you want an array of the type SomeObject, you could use:
hashMap.keySet().toArray(new SomeObject[0]);
To guarantee the correct order for each array of Keys and Values, use this (the other answers use individual Sets which offer no guarantee as to order.
Map&String, Object& map = new HashMap&String, Object&();
String[] keys = new String[map.size()];
Object[] values = new Object[map.size()];
int index = 0;
for (Map.Entry&String, Object& mapEntry : map.entrySet()) {
keys[index] = mapEntry.getKey();
values[index] = mapEntry.getValue();
2,89911538
An alternative to CrackerJacks suggestion, if you want the HashMap to maintain order you could consider using a LinkedHashMap instead. As far as im aware it's functionality is identical to a HashMap but it is FIFO so it maintains the order in which items were added.
2,40112030
Map&String, String& map = new HashMap&String, String&();
map.put("key1", "value1");
map.put("key2", "value2");
Object[][] twoDarray = new String[map.size()][2];
Object[] keys = map.keySet().toArray();
Object[] values = map.values().toArray();
for (int row = 0; row & twoDarray. row++) {
twoDarray[row][0] = keys[row];
twoDarray[row][1] = values[row];
for (int i = 0; i & twoDarray. i++) {
for (int j = 0; j & twoDarray[i]. j++) {
System.out.println(twoDarray[i][j]);
77.6k16148248
To Get in One Dimension Array.
String[] arr1 = new String[hashmap.size()];
String[] arr2 = new String[hashmap.size()];
Set entries = hashmap.entrySet();
Iterator entriesIterator = entries.iterator();
int i = 0;
while(entriesIterator.hasNext()){
Map.Entry mapping = (Map.Entry) entriesIterator.next();
arr1[i] = mapping.getKey().toString();
arr2[i] = mapping.getValue().toString();
To Get in two Dimension Array.
String[][] arr = new String[hashmap.size()][2];
Set entries = hashmap.entrySet();
Iterator entriesIterator = entries.iterator();
int i = 0;
while(entriesIterator.hasNext()){
Map.Entry mapping = (Map.Entry) entriesIterator.next();
arr[i][0] = mapping.getKey().toString();
arr[i][1] = mapping.getValue().toString();
You may try this too.
public static String[][] getArrayFromHash(Hashtable&String,String& data){
String[][] str =
Object[] keys = data.keySet().toArray();
Object[] values = data.values().toArray();
str = new String[keys.length][values.length];
for(int i=0;i&keys.i++) {
str[0][i] = (String)keys[i];
str[1][i] = (String)values[i];
Here I am using String as return type. You may change it to required return type by you.
I used almost the same as @kmccoy, but instead of a keySet() I did this
hashMap.values().toArray(new MyObject[0]);
Your Answer
Sign up or
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Post as a guest
By posting your answer, you agree to the
Not the answer you're looking for?
Browse other questions tagged
Stack Overflow works best with JavaScript enabled