关于排序

本帖最后由赵庆礼于 2012-6-28 14:30 编辑

{seqno=0, amt=1000, id=BOCC, date=20101010}
{seqno=1, amt=1000, id=RRRR, date=20101010}
{seqno=2, amt=1000, id=BOCC, date=20101009}
{seqno=3, amt=1000, id=BOCC, date=20101008}
{seqno=4, amt=1000, id=RRRR, date=20101013}
{seqno=5, amt=2000, id=RRRR, date=20101013}
.......
........

需求:
整体为一个vector
每行为一个map

按id来分组,相同id中取日期最小的金额
求加和
有什么好的办法吗?

我的解法:
循环整个vector,distinct(id),
再循环id,内循环整个Vector,取出各个相同的id分组多个Vector,
再单独循环每一个vector取出各组中日期最小的...
很麻烦...

周朋飞 · 周朋飞

我写一个你试试吧
public class T001 {
public static void main(String[] args) {
Vector v = getV();
Map amtMap = getMixDateAmt(v);

double sum = 0;
Iterator iter = amtMap.keySet().iterator();
while(iter.hasNext()) {
sum += Double.parseDouble((String)amtMap.get(iter.next()));
}
}

private static Map getMixDateAmt(Vector v) {
Map dateMap = new HashMap();
Map amtMap = new HashMap();

String id1 = (String)((Map)v.get(0)).get("id");
String date1 = (String)((Map)v.get(0)).get("date");
String amt1 = (String)((Map)v.get(0)).get("amt");

dateMap.put(id1, date1);
amtMap.put(id1, amt1);

for(int i = 0; i < v.size(); i++) {
Map m = (Map)v.get(i);
String id2 = (String)m.get("id");
String date2 = (String)m.get("date");
String amt2 = (String)m.get("amt");

if(!dateMap.containsKey(id2)) {
dateMap.put(id2, date2);
amtMap.put(id2, amt2);
} else {
if(date2.compareTo((String)dateMap.get(id2)) < 0) {
dateMap.put(id2, date2);
amtMap.put(id2, amt2);
}
}
}
return amtMap;
}

public static ArrayList getDistinctId(Vector v) {
String id1 = (String)((Map)v.get(0)).get("id");
ArrayList list = new ArrayList();

list.add(id1);
for(int i = 0; i < v.size(); i++) {
Map m2 = (Map)v.get(i);
String id2 = (String)m2.get("id");
if(!id1.equals(id2) && !list.contains(id2)) {
list.add(id2);
}
}
System.out.println(list.toString());
return list;
}
}

黑马刘杰 · 黑马刘杰

本帖最后由 Jackey 于 2012-6-28 12:50 编辑

好复杂！

周朋飞 · 周朋飞

这是我调通了的程序你要是哪点不理解可以发给我或者自己先debug一下

package testDemo;

import java.util.ArrayList;
import java.util.HashMap;
import java.util.Iterator;
import java.util.Map;
import java.util.Vector;

public class Test {
public static void main(String[] args) {
Vector v = getV();

Map amtMap = getMixDateAmt(v);

double sum = 0;
Iterator iter = amtMap.keySet().iterator();
while (iter.hasNext()) {
sum += Double.parseDouble((String) amtMap.get(iter.next()));
System.out.println(sum);
}
System.out.println(sum);
}

private static Vector getV() {
// TODO Auto-generated method stub
Vector vector = new Vector();
Map<String, String> map1 = new HashMap<String, String>();
Map<String, String> map2 = new HashMap<String, String>();
Map<String, String> map3 = new HashMap<String, String>();
Map<String, String> map4 = new HashMap<String, String>();
Map<String, String> map5 = new HashMap<String, String>();
map1.put("seqno", "0");
map1.put("amt", "1000");
map1.put("id", "1");
map1.put("date", " 20101010");

map2.put("seqno", "1");
map2.put("amt", "1000");
map2.put("id", "2");
map2.put("date", " 20101010");

map3.put("seqno", "2");
map3.put("amt", "1000");
map3.put("id", "1");
map3.put("date", " 20101009");

map4.put("seqno", "3");
map4.put("amt", "1000");
map4.put("id", "1");
map4.put("date", " 20101008");

map5.put("seqno", " 1");
map5.put("amt", "1000");
map5.put("id", "2");
map5.put("date", " 20101013");
vector.add(map1);
vector.add(map2);
vector.add(map3);
vector.add(map4);
vector.add(map5);

return vector;
}

private static Map getMixDateAmt(Vector v) {
Map dateMap = new HashMap();
Map amtMap = new HashMap();

String id1 = (String) ((Map) v.get(0)).get("id");
String date1 = (String) ((Map) v.get(0)).get("date");
String amt1 = (String) ((Map) v.get(0)).get("amt");

dateMap.put(id1, date1);
amtMap.put(id1, amt1);

for (int i = 0; i < v.size(); i++) {
Map m = (Map) v.get(i);
String id2 = (String) m.get("id");
String date2 = (String) m.get("date");
String amt2 = (String) m.get("amt");

if (!dateMap.containsKey(id2)) {
dateMap.put(id2, date2);
amtMap.put(id2, amt2);
} else {
if (date2.compareTo((String) dateMap.get(id2)) < 0) {
dateMap.put(id2, date2);
amtMap.put(id2, amt2);
}
}
}
return amtMap;
}

public static ArrayList getDistinctId(Vector v) {
String id1 = (String) ((Map) v.get(0)).get("id");
ArrayList list = new ArrayList();

list.add(id1);
for (int i = 0; i < v.size(); i++) {
Map m2 = (Map) v.get(i);
String id2 = (String) m2.get("id");
if (!id1.equals(id2) && !list.contains(id2)) {
list.add(id2);
}
}
System.out.println(list.toString());
return list;
}
}

周朋飞 · 周朋飞

这是我调通了的程序你要是哪点不理解可以发给我或者自己先debug一下

package testDemo;

import java.util.ArrayList;
import java.util.HashMap;
import java.util.Iterator;
import java.util.Map;
import java.util.Vector;

public class Test {
public static void main(String[] args) {
Vector v = getV();

Map amtMap = getMixDateAmt(v);

double sum = 0;
Iterator iter = amtMap.keySet().iterator();
while (iter.hasNext()) {
sum += Double.parseDouble((String) amtMap.get(iter.next()));
System.out.println(sum);
}
System.out.println(sum);
}

private static Vector getV() {
// TODO Auto-generated method stub
Vector vector = new Vector();
Map<String, String> map1 = new HashMap<String, String>();
Map<String, String> map2 = new HashMap<String, String>();
Map<String, String> map3 = new HashMap<String, String>();
Map<String, String> map4 = new HashMap<String, String>();
Map<String, String> map5 = new HashMap<String, String>();
map1.put("seqno", "0");
map1.put("amt", "1000");
map1.put("id", "1");
map1.put("date", " 20101010");

map2.put("seqno", "1");
map2.put("amt", "1000");
map2.put("id", "2");
map2.put("date", " 20101010");

map3.put("seqno", "2");
map3.put("amt", "1000");
map3.put("id", "1");
map3.put("date", " 20101009");

map4.put("seqno", "3");
map4.put("amt", "1000");
map4.put("id", "1");
map4.put("date", " 20101008");

map5.put("seqno", " 1");
map5.put("amt", "1000");
map5.put("id", "2");
map5.put("date", " 20101013");
vector.add(map1);
vector.add(map2);
vector.add(map3);
vector.add(map4);
vector.add(map5);

return vector;
}

private static Map getMixDateAmt(Vector v) {
Map dateMap = new HashMap();
Map amtMap = new HashMap();

String id1 = (String) ((Map) v.get(0)).get("id");
String date1 = (String) ((Map) v.get(0)).get("date");
String amt1 = (String) ((Map) v.get(0)).get("amt");

dateMap.put(id1, date1);
amtMap.put(id1, amt1);

for (int i = 0; i < v.size(); i++) {
Map m = (Map) v.get(i);
String id2 = (String) m.get("id");
String date2 = (String) m.get("date");
String amt2 = (String) m.get("amt");

if (!dateMap.containsKey(id2)) {
dateMap.put(id2, date2);
amtMap.put(id2, amt2);
} else {
if (date2.compareTo((String) dateMap.get(id2)) < 0) {
dateMap.put(id2, date2);
amtMap.put(id2, amt2);
}
}
}
return amtMap;
}

public static ArrayList getDistinctId(Vector v) {
String id1 = (String) ((Map) v.get(0)).get("id");
ArrayList list = new ArrayList();

list.add(id1);
for (int i = 0; i < v.size(); i++) {
Map m2 = (Map) v.get(i);
String id2 = (String) m2.get("id");
if (!id1.equals(id2) && !list.contains(id2)) {
list.add(id2);
}
}
System.out.println(list.toString());
return list;
}
}

李天甲 · 李天甲

晕了大哥们,提交问题的时候请把条件写的清楚一点呵呵,请尽量改一下名字,在网上搜不出来为妙
原帖请参见http://topic.csdn.net/u/20101214/13/eb1561c0-4f4d-403e-a4d5-56186735b9ae.html

这个题目解起来不难,
跟提问题的人的思路差不多

先循环整个vector,distinct(id),注意这里产生的是一个只有唯一ID的list

再循环id,(这个?呵呵,上面解题的同学你没用上上面的参数呵呵)
内循环整个Vector,取出各个相同的id分组多个Vector,

再单独循环每一个vector取出各组中日期最小的...

基本就是这样了

赵庆礼 · 赵庆礼

周朋飞发表于 2012-6-28 12:15
我写一个你试试吧
public class T001 {
public static void main(String[] args) {

周，我看明白了，谢谢

周朋飞 · 周朋飞

既然写了就多写点吧，其实对Map排序有两种方法，一种是题中所提用keySet方法，另一种方法更为高效就是通过entrySet来做
double sum = 0;
Set<Map.Entry<String, String>> entrys = amtMap.entrySet();
for(Map.Entry entry:entrys){

sum = sum +Double.parseDouble((String)entry.getValue());
}

周朋飞 · 周朋飞

既然写了就多写点吧，其实对Map排序有两种方法，一种是题中所提用keySet方法，另一种方法更为高效就是通过entrySet来做
double sum = 0;
Set<Map.Entry<String, String>> entrys = amtMap.entrySet();
for(Map.Entry entry:entrys){

sum = sum +Double.parseDouble((String)entry.getValue());
}
最后那个方法可有可无，用来对id进行分组看你个人需要

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