// 统计
public ArrayList summaryList(ArrayList arrayList, String Name) { ArrayList resultList = new ArrayList();
while (arrayList.size() > 1) {
HashMap hm1 = (HashMap) arrayList.get(0);
String totalAmount = "0";
String totalconvertAmount = "0";
for (int ii = 1; arrayList.size() > ii;) {
HashMap hm2 = (HashMap) arrayList.get(ii);
String key1 = (String) hm1.get(Name);
String key2 = (String) hm2.get(Name);
String cur1 = (String) hm1.get("Currency");
String cur2 = (String) hm2.get("Currency");
if (key1.hashCode() == key2.hashCode() && cur1.hashCode() == cur2.hashCode()) {
BigDecimal temp1 = new BigDecimal(0);
BigDecimal temp2 = new BigDecimal(0);
String sum1 = (String) hm1.get("Amount");
String sum2 = (String) hm2.get("Amount");
String sum3 = (String) hm1.get("ConvertRMB");
String sum4 = (String) hm2.get("ConvertRMB");
BigDecimal a = new BigDecimal(sum1);
BigDecimal b = new BigDecimal(sum2);
BigDecimal c = new BigDecimal(sum3);
BigDecimal d = new BigDecimal(sum4);
temp1 = temp1.add(a);
temp1 = temp1.add(b);
temp2 = temp2.add(c);
temp2 = temp2.add(d);
totalAmount = temp1.toString();
totalconvertAmount = temp2.toString();
hm1.put("Amount", totalAmount);
hm1.put("ConvertRMB", totalconvertAmount);
arrayList.remove(hm2);//与arrayList.remove(ii);的区别?
continue;
}
ii++;
}
resultList.add(hm1);
arrayList.remove(hm1);//与arrayList.remove(0);的区别? }
if (arrayList.size() == 1) {
resultList.add(arrayList.get(0));
}
return resultList;
}
public ArrayList summaryList(ArrayList arrayList, String Name) { ArrayList resultList = new ArrayList();
while (arrayList.size() > 1) {
HashMap hm1 = (HashMap) arrayList.get(0);
String totalAmount = "0";
String totalconvertAmount = "0";
for (int ii = 1; arrayList.size() > ii;) {
HashMap hm2 = (HashMap) arrayList.get(ii);
String key1 = (String) hm1.get(Name);
String key2 = (String) hm2.get(Name);
String cur1 = (String) hm1.get("Currency");
String cur2 = (String) hm2.get("Currency");
if (key1.hashCode() == key2.hashCode() && cur1.hashCode() == cur2.hashCode()) {
BigDecimal temp1 = new BigDecimal(0);
BigDecimal temp2 = new BigDecimal(0);
String sum1 = (String) hm1.get("Amount");
String sum2 = (String) hm2.get("Amount");
String sum3 = (String) hm1.get("ConvertRMB");
String sum4 = (String) hm2.get("ConvertRMB");
BigDecimal a = new BigDecimal(sum1);
BigDecimal b = new BigDecimal(sum2);
BigDecimal c = new BigDecimal(sum3);
BigDecimal d = new BigDecimal(sum4);
temp1 = temp1.add(a);
temp1 = temp1.add(b);
temp2 = temp2.add(c);
temp2 = temp2.add(d);
totalAmount = temp1.toString();
totalconvertAmount = temp2.toString();
hm1.put("Amount", totalAmount);
hm1.put("ConvertRMB", totalconvertAmount);
arrayList.remove(hm2);//与arrayList.remove(ii);的区别?
continue;
}
ii++;
}
resultList.add(hm1);
arrayList.remove(hm1);//与arrayList.remove(0);的区别? }
if (arrayList.size() == 1) {
resultList.add(arrayList.get(0));
}
return resultList;
}
QueryTotalAsset a = new QueryTotalAsset();
ArrayList list = new ArrayList();
HashMap map1 = new HashMap();
map1.put("Currency","01");
map1.put("CDType","HQ");
map1.put("Amount","0.0");
map1.put("ConvertRMB","0.0");
HashMap map2 = new HashMap();
map2.put("Currency","01");
map2.put("CDType","HQ");
map2.put("Amount","2015.44");
map2.put("ConvertRMB","2015.44");
HashMap map3 = new HashMap();
map3.put("Currency","01");
map3.put("CDType","HQ");
map3.put("Amount","67.69");
map3.put("ConvertRMB","67.69"); list.add(map1);
list.add(map2);
list.add(map3);
//区别
// list.add(map2);
// list.add(map1);
// list.add(map3);
System.out.println(a.summaryList(list,"CDType"));
}
=====================================================================================
remove(object):
[{Amount=2083.13, ConvertRMB=2083.13, CDType=HQ, Currency=01}, {Amount=2015.44, ConvertRMB=2015.44, CDType=HQ, Currency=01}]
================================================
remove(int):[{Amount=2083.13, ConvertRMB=2083.13, CDType=HQ, Currency=01}]
remove(int)是移除指定位置的对象
我也奇怪,既然前面用get(int i)方式,后面就用remove(int i)呗,为什么换成remove(Object obj)
如果用remove(int i)两次执行结果是一致的!!!我想坛子里明白的人帮解释下,是不是
原因是他一直把相加总和存在第一个map里面,他原来设想是把总数相加之后,删除掉当前map,但是当第二个map出现的时候,总和相加了,但是数值确是hm1和hm2相同了,造成了这个list中有两个相同内容的map,然后remove(hm2),实际删除的却是hm1,以为这时hm1.equals(hm2)是true;
后一次循环正常进行,并且结束。
但是这时list里面剩下了一条原先map2的数据,而不是他想要的map1的数据,所以造成了remove失败。所以这种题目应该还是以remove(int i)的方式比较保险