今天開發的時候遇到個小問題，解決后拿出來曬曬，看看大家還有沒有更好的辦法，呵呵！
要求：兩個字符串數組集合，A數組兩個元素code和name，B數組兩個元素code和number，需要把B數組中沒有A數組的code進行補缺并number置零，并且要按A數組code的排序方式。
舉例：
List aList = new ArrayList();
aList.add(new String[]{"1","name1"});
aList.add(new String[]{"2","name2"});
aList.add(new String[]{"3","name3"});
List bList1 = new ArrayList();
bList1.add(new String[]{"1","5"});
bList1.add(new String[]{"2","5"});
List bList2 = new ArrayList();
bList2.add(new String[]{"3","5"});
解答：
public List handle(List bList, List aList) {
for (Object object : aList) {
String[] a = (String[]) object;
boolean flag = false;
for (Object object1 : bList) {
String[] b = (String[]) object1;
if(b[0].equals(a[0])) {
flag = true;
break;
}
}
if(!flag) {
String[] b = (String[]) bList.get(0);
String[] newA = new String[b.length];
newA[0] = a[0];
newA[1] = "0";
bList.add(newA);
}
}
return bList;
}
為了排序還需要加一個排序類
class exComparator implements Comparator {
public int compare(Object arg0, Object arg1) {
String[] vo1 = (String[])arg0;
String[] vo2 = (String[])arg1;
int result = 0;
if(Integer.parseInt(vo1[0])<Integer.parseInt(vo2[0])) {
result = -1;
} else if(Integer.parseInt(vo1[0])>Integer.parseInt(vo2[0])) {
result = 1;
} else {
result = 0;
}
return result;
}
}
驗證:
List example = handle(bList1,aList);
Collections.sort(example, new exComparator());
for (Object object : example) {
String[] ss = (String[]) object;
System.out.println("========"+ss[0]+"=======" + ss[1]);
}
example = handle(bList2,aList);
Collections.sort(example, new exComparator());
for (Object object : example) {
String[] ss = (String[]) object;
System.out.println("========"+ss[0]+"=======" + ss[1]);
}
結果：
========1=======5
========2=======5
========3=======0
========1=======0
========2=======0
========3=======5
over............