首页 > 代码库 > [Java]HashMap的两种排序方式

[Java]HashMap的两种排序方式

2024-07-06 08:22:05   236人阅读

先将 Map 中的 key 和 value 全部取出来封装成 JavaBea 数组，再将这个数组排序，
排序完成后，重新写回 Map 中，写回时采用 LinkedHashMap 可以保证迭代的顺序。

下面的代码可以参考一下：

?
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
import java.util.ArrayList;
import java.util.Collections;
import java.util.Comparator;
import java.util.HashMap;
import java.util.Iterator;
import java.util.LinkedHashMap;
import java.util.List;
import java.util.Map;
 
public class Test {
 
    public static void main(String[] args) {
        Map<String, Integer> map = new HashMap<String, Integer>();
        map.put("ee", 3);
        map.put("b", 1);
        map.put("d", 2);
        map.put("eee", 3);
        map.put("A", 1);
        map.put("K", 2);
        map.put("ade", 1);
        map.put("c", 2);
        map.put("aee", 3);
        map.put("a", 1);
        map.put("faed", 2);
        map.put("bdd", 1);
        map.put("qec", 2);
        map.put("eade", 3);
        map.put("Aadf", 1);
        map.put("Kqe", 2);
 
        Map<String, Integer> sortMap = new Test().sortMap(map);
 
        for(Map.Entry<String, Integer> entry : sortMap.entrySet()) {
            System.out.println(entry.getKey() + " --> " + entry.getValue());
        }
    }
 
    public <K, V extends Number> Map<String, V> sortMap(Map<String, V> map) {
        class MyMap<M, N> {
            private M key;
            private N value;
            private M getKey() {
                return key;
            }
            private void setKey(M key) {
                this.key = key;
            }
            private N getValue() {
                return value;
            }
            private void setValue(N value) {
                this.value = http://www.mamicode.com/value;
            }
        }
 
        List<MyMap<String, V>> list = new ArrayList<MyMap<String, V>>();
        for (Iterator<String> i = map.keySet().iterator(); i.hasNext(); ) {
            MyMap<String, V> my = new MyMap<String, V>();
            String key = i.next();
            my.setKey(key);
            my.setValue(map.get(key));
            list.add(my);
        }
 
        Collections.sort(list, new Comparator<MyMap<String, V>>() {
            public int compare(MyMap<String, V> o1, MyMap<String, V> o2) {
                if(o1.getValue() == o2.getValue()) {
                    return o1.getKey().compareTo(o2.getKey());
                }else{
                    return (int)(o1.getValue().doubleValue() - o2.getValue().doubleValue());
                }
            }
        });
 
        Map<String, V> sortMap = new LinkedHashMap<String, V>();
        for(int i = 0, k = list.size(); i < k; i++) {
            MyMap<String, V> my = list.get(i);
            sortMap.put(my.getKey(), my.getValue());
        }
        return sortMap;
    }
}

　　

Map<String, Integer> map = new HashMap<String, Integer>();
map.put("d", 2);
map.put("c", 1);
map.put("b", 1);
map.put("a", 3);

List<Map.Entry<String, Integer>> infoIds =
new ArrayList<Map.Entry<String, Integer>>(map.entrySet());

//排序前
for (int i = 0; i < infoIds.size(); i++) {
String id = infoIds.get(i).toString();
System.out.println(id);
}
//d 2
//c 1
//b 1
//a 3

//排序
Collections.sort(infoIds, new Comparator<Map.Entry<String, Integer>>() {
public int compare(Map.Entry<String, Integer> o1, Map.Entry<String, Integer> o2) {
//return (o2.getValue() - o1.getValue());
return (o1.getKey()).toString().compareTo(o2.getKey());
}
});

//排序后
for (int i = 0; i < infoIds.size(); i++) {
String id = infoIds.get(i).toString();
System.out.println(id);
}
//根据key排序
//a 3
//b 1
//c 1
//d 2
//根据value排序
//a 3
//d 2
//b 1
//c 1

 

参考：   http://bbs.csdn.net/topics/230054066

　　　　http://www.cnblogs.com/lovebread/archive/2009/11/23/1609121.html


联系
我们