以下代码将重复的名称分成 1 列,并将与名称关联的数字总和分成第二列。
喜欢:
Nokia 21
Blackberry 3
Nimbus 30
来自程序中给出的数组。
<小时/>我想知道包含这些条目的数组的最终长度。在这种情况下 3. 我如何计算?
package keylogger;
import java.util.ArrayList;
import java.util.List;
public class ArrayTester {
private static int finalLength = 0;
private static String Name[][];
private static String data[][] = {
{"Nokia" , "7"},
{"Blackberry" ,"1"},
{"Nimbus","10"},
{"Nokia" , "7"},
{"Blackberry" , "1"},
{"Nimbus","10"},
{"Nokia" , "7"},
{"Blackberry" , "1"},
{"Nimbus","10"}
};
public void calculator() {
Name = new String[data.length][2];
List<String> marked = new ArrayList<String>();
try {
for(int i=0;i<data.length;i++) {
Name[i][0] = data[i][0];
Name[i][1] = data[i][1];
String name = data[i][0];
if(marked.contains(name)) {
continue;
}
marked.add(name);
int k = i + 1;
int v = k;
for (int j = 0; j < data.length - v; j++) {
String s = data[k][0];
if(Name[i][0].equalsIgnoreCase(s)) {
Name[i][0] = s;
Integer z = Integer.parseInt(Name[i][1]) + Integer.parseInt(data[k][1]);
Name[i][1] = z.toString();
}
k++;
}
}
}catch(Exception exc) {
exc.printStackTrace();
}
}
public static void main(String args[]) {
ArrayTester o = new ArrayTester();
o.calculator();
for(String s[] : Name) {
for(String x : s) {
System.out.println(x);
}
}
}
}
最佳答案
像往常一样,“问题”是糟糕的编码。如果编写得当,整个程序可以减少到只有 3 行代码(如果包括定义数组和打印输出,则为 5 行):
public static void main(String[] args) {
String data[][] = {{"Nokia", "7"}, {"Blackberry", "1"}, {"Nimbus", "10"},
{"Nokia", "7"}, {"Blackberry", "1"}, {"Nimbus", "10"}, {"Nokia", "7"},
{"Blackberry", "1"}, {"Nimbus", "10"}, {"Zebra", "78"}};
HashMap<String, Integer> totals = new HashMap<String, Integer>();
for (String[] datum : data)
totals.put(datum[0], new Integer(datum[1]) + (totals.containsKey(datum[0]) ? totals.get(datum[0]) : 0));
System.out.println("There are " + totals.size() + " brands: " + totals);
}
输出:
There are 4 brands: {Nimbus=30, Zebra=78, Nokia=21, Blackberry=3}
关于java - 计算最终长度,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/10754764/