通过以下代码，我希望将所有以相同名称开头的字母组合到一个数组列中，并在第二列中保留与字母相关的数字的总和。
例如:
array_1 = { {"bat","1"},
{"rat","2"},
{"ball","3"},
{"run","4"},
{"lizard","5"}
}
into array_2 = { {"b","4"},
{"r","6"},
{"l",5}
}
下面的代码给出了一半正确的结果。问题是，当它到达 ball 时，它会再次添加以 b 开头的下一个字母并将其存储为单独的值。问题是第 42 行。我已经标记了这一点。我应该如何进行检查，确保添加后不会添加字母表的数字。
package keylogger;
import java.util.Arrays;
public class ArrayTester {
private static int finalLength=0;
private static String firstAlphabet[][];
private String data[][] = {
{"Nokia" , "7"},
{"Blackberry" , "1"},
{"Nimbus","10"},
{"Apple","19"},
{"Ami","21"},
{"Bug","35"},
{"Native","200"},
{"zebra","100"},
{"Nine","9"}
};
public void calculator() {
try {
// traverse the whole array
firstAlphabet = new String[data.length][data.length]; // set the length of firstAlphabet array
for(int i=0;i<data.length;i++) {
firstAlphabet[i][0] = data[i][0].substring( 0, 1); // get the first alphabet
firstAlphabet[i][1] = data[i][1];
int k = i+1;
int v = k;
int t=0;
for(int j=0;j<data.length-v;j++) {
System.out.println("Inner for loop" + j);
String s = data[k][0];
// line 42:
if(firstAlphabet[i][0].compareToIgnoreCase(s.substring(0, 1))==0) {
System.out.println("If statement");
firstAlphabet[i][0] = s.substring(0, 1);
Integer z = Integer.parseInt(data[k][1]) + Integer.parseInt(firstAlphabet[i][1]);
firstAlphabet[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[] : firstAlphabet) {
for(String x : s) {
System.out.println(x);
}
}
}
}
输出
Inner for loop0
Inner for loop1
If statement
Inner for loop2
Inner for loop3
Inner for loop4
Inner for loop5
If statement
Inner for loop6
Inner for loop7
If statement
Inner for loop0
Inner for loop1
Inner for loop2
Inner for loop3
If statement
Inner for loop4
Inner for loop5
Inner for loop6
Inner for loop0
Inner for loop1
Inner for loop2
Inner for loop3
If statement
Inner for loop4
Inner for loop5
If statement
Inner for loop0
If statement
Inner for loop1
Inner for loop2
Inner for loop3
Inner for loop4
Inner for loop0
Inner for loop1
Inner for loop2
Inner for loop3
Inner for loop0
Inner for loop1
Inner for loop2
Inner for loop0
Inner for loop1
If statement
Inner for loop0
N
226
null
null
null
null
null
null
null
B
36
null
null
null
null
null
null
null
N
219
null
null
null
null
null
null
null
A
40
null
null
null
null
null
null
null
A
21
null
null
null
null
null
null
null
B
35
null
null
null
null
null
null
null
N
209
null
null
null
null
null
null
null
z
100
null
null
null
null
null
null
null
N
9
null
null
null
null
null
null
null
如果我们注意到与字母相关的总和是正确的。唯一的问题是重复。例如，N=219，即 200+19
最佳答案
重复是因为您没有将已选择的字母标记为脏。
因此，在第一个循环中，N 本身已获得最终计数，但根据您当前的逻辑，当 Nimbus 进行迭代时，您将完成整个处理，因为您尚未将 N 标记为脏。
public void calculator() {
List<String> marked = new ArrayList<String>();
try {
// traverse the whole array
firstAlphabet = new String[data.length][2]; // set the length of first Alphabet array
for (int i = 0; i < data.length; i++) {
String firstLetter = data[i][0].substring(0, 1);
if(marked.contains(firstLetter)){
continue;
}
marked.add(firstLetter);
firstAlphabet[i][0] = firstLetter; // get the first alphabet
firstAlphabet[i][1] = data[i][1];
int k = i + 1;
int v = k;
int t = 0;
for (int j = 0; j < data.length - v; j++) {
System.out.println("Inner for loop" + j);
String s = data[k][0];
if (firstAlphabet[i][0].equalsIgnoreCase(s.substring(0,
1))) { // line 42
System.out.println("If statement");
firstAlphabet[i][0] = s.substring(0, 1);
Integer z = Integer.parseInt(data[k][1])
+ Integer.parseInt(firstAlphabet[i][1]);
firstAlphabet[i][1] = z.toString();
}
k++;
}
}
} catch (Exception exc) {
exc.printStackTrace();
}
}
关于java - 强行检查 : don't proceed if there is already a string in column number 0，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/10752857/