本文介绍了无法按升序排序顺序列表的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!
问题描述
地图<字符串,字符串>图;
清单<地图<字符串,字符串>>名单=新的ArrayList<地图<字符串,字符串>>();
/////的OnCreate .............
功能1(){
地图=新TreeMap的<字符串,字符串>();
map.put(ID,ID);
map.put(入账金额,金额);list.add(地图);的System.out.println(名单);
}
id的输入值= 1,3,5,57,80
有关量的输入值= 100,500,200,10,10000
无法按升序排序多少的顺序列表。它仍然显示了它被插入的顺序。
我该如何解决这个问题?我AP preciate任何帮助。先谢谢了。
预计产量:量的升序:
AMT = 10 ID = 4
AMT = 100的id = 1
AMT = 200的id = 3
AMT = 500 ID = 2
AMT = 10000 ID = 5
解决方案
假设这是你输入
地图<字符串,字符串>图;
清单<地图<字符串,字符串>>名单=新的ArrayList<地图<字符串,字符串>>();
地图=新TreeMap的<字符串,字符串>();
map.put(ID,1);
map.put(量,100);
list.add(地图);
地图=新TreeMap的<字符串,字符串>();
map.put(ID,2);
map.put(量,500);
list.add(地图);
地图=新TreeMap的<字符串,字符串>();
map.put(ID,3);
map.put(量,200);
list.add(地图);
地图=新TreeMap的<字符串,字符串>();
map.put(ID,4);
map.put(量,10);
list.add(地图);
地图=新TreeMap的<字符串,字符串>();
map.put(ID,5);
map.put(额,10000);
list.add(地图);
下面是您的排序code
Collections.sort(名单,新的比较<地图<字符串,字符串>>(){ @覆盖
公众诠释比较(地图<字符串,字符串> 01,地图<字符串,字符串> O2){
字符串值1 = o1.get(额);
字符串值2 = o2.get(额);
返回的Integer.parseInt(值)-Integer.parseInt(值2);
}
}); 对于(地图<字符串,字符串> MAP1:名单){
字符串ID = map1.get(ID);
串金额= map1.get(额);
的System.out.println(金额=量+ +,+的id =+ ID);
}
输出
额= 10,ID = 4
量= 100,ID = 1
量= 200,ID = 3
量= 500,ID = 2
量= 10000,ID = 5
更新
替换返回的Integer.parseInt(值)-Integer.parseInt(值2);
通过以下code。如果值是小数
返回Double.valueOf(值).compareTo(Double.valueOf(值2));
Map<String, String> map ;
List<Map<String, String>> list = new ArrayList<Map<String, String>>();
/////OnCreate.............
function1(){
map = new TreeMap<String, String>();
map.put("id", "id");
map.put("amont", "amount");
list.add(map);
System.out.println(list);
}
input values for id=1,3,5,57,80
input values for amount=100,500,200,10,10000
Unable to sort the list by ascending order of amount. It still shows in the order it was inserted.
How do I fix this? I appreciate any help. Thanks in advance.
Expected output: Ascending order of amount:
amt=10 id=4
amt=100 id=1
amt=200 id=3
amt=500 id=2
amt=10000 id=5
解决方案
Assuming this is your input
Map<String, String> map ;
List<Map<String, String>> list = new ArrayList<Map<String, String>>();
map = new TreeMap<String, String>();
map.put("id","1");
map.put("amount","100");
list.add(map);
map = new TreeMap<String, String>();
map.put("id","2");
map.put("amount","500");
list.add(map);
map = new TreeMap<String, String>();
map.put("id","3");
map.put("amount","200");
list.add(map);
map = new TreeMap<String, String>();
map.put("id","4");
map.put("amount","10");
list.add(map);
map = new TreeMap<String, String>();
map.put("id","5");
map.put("amount","10000");
list.add(map);
Here is your sorting code
Collections.sort(list, new Comparator<Map<String, String>>() {
@Override
public int compare(Map<String, String> o1, Map<String, String> o2) {
String value1 = o1.get("amount");
String value2 = o2.get("amount");
return Integer.parseInt(value1)-Integer.parseInt(value2);
}
});
for (Map<String, String> map1 : list) {
String id = map1.get("id");
String amount = map1.get("amount");
System.out.println("amount= "+amount + " , " +"id = "+id);
}
Output
amount= 10 , id = 4
amount= 100 , id = 1
amount= 200 , id = 3
amount= 500 , id = 2
amount= 10000 , id = 5
update
Replace return Integer.parseInt(value1)-Integer.parseInt(value2);
with the following code if the values are decimal.
return Double.valueOf(value1).compareTo(Double.valueOf(value2));
这篇关于无法按升序排序顺序列表的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持!