Is it possible to use Integer.parse for String array of "mixed" types?

Question

I have a String array

String[] list = new String[]{"2","A","B","1","C","3","D","E","F"}

Please Note that the first element of array is 2. I would have done -

int n = Integer.parseInt(list[0]);

Problem is, I need a loop to go through the array. I cant seem to parse it without giving me a NumberFormatException error. Also, notice that if the number is a 2, it will arrange the next 2 elements (might be random) and then a space.

1: A

2: B

3: -empty-

4: D

5: E

6: F

7: -empty-

8: C

possible duplicate of [What's the best way to check to see if a String represents an integer in Java?](http://stackoverflow.com/questions/237159/whats-the-best-way-to-check-to-see-if-a-string-represents-an-integer-in-java) — ChiefTwoPencils, Sep 17 '15 at 05:16
Blindly you can't do it. Why would you even do it ? As your problem statement says, converting whole to array to integer doesn't seem your purpose. — SacJn, Sep 17 '15 at 05:17
Do you mean to say that the number in the array specifies the number of elements that are string following in the array and retrieve the string elements only? — Blip, Sep 17 '15 at 05:26
@aurora_titanium please post a reply to my previous comment as a new comment if you require help — Blip, Sep 17 '15 at 05:43
You don't have to face NFE if you first read digits then read digits many characters without converting them to integer. Control is in your hands so you can definitely avoid NFE — SacJn, Sep 17 '15 at 05:57

score 0 · Answer 1 · edited May 23 '17 at 12:22

0

Iterate through the array and check whether the selected character is a digit or not. If it's a digit use parseInt.

Here's a tutorial on how to identify digits. What is the best way to tell if a character is a letter or number in Java without using regexes?

Good luck.

edited May 23 '17 at 12:22

Community

1
1

answered Sep 17 '15 at 05:18

Nimila Hiranya

4,842
10
35
52

score 0 · Answer 2 · answered Sep 17 '15 at 05:22

Use a try/catch statement in your loop when you iterate over the array.

Here's a short example:

for(int i = 0; i < list.length; i++) {
    try {
        int n = Integer.parseInt(list[i]);
        // do whatever you need to do with n here
    }
    catch (NumberFormatException nfex) {
        // do something else with the current element if need be
    }
}

score 0 · Answer 3 · answered Sep 17 '15 at 05:24

0

for(String hexNbr : list) 
{
   System.out.println("Next number is " + Integer.parseInt(hexNbr, 16);
}

Your input is hex, so you have to use parseInt(String, radix).

answered Sep 17 '15 at 05:24

Brad

2,261
3
22
32

score 0 · Accepted Answer · answered Sep 17 '15 at 06:08

Algorithm is simple enough.

Parse number
Shift N values 1 position to the left
Assign empty value to vacant space
Advance past block, and repeat

Progress is a follows:

Start: "2","A","B","1","C","3","D","E","F"
N = 2
Shift: "A","B","B","1","C","3","D","E","F"
Empty: "A","B","-","1","C","3","D","E","F"
N = 1
Shift: "A","B","-","C","C","3","D","E","F"
Empty: "A","B","-","C","-","3","D","E","F"
N = 3
Shift: "A","B","-","C","-","D","E","F","F"
Empty: "A","B","-","C","-","D","E","F","-"

Code

String[] list = {"2","A","B","1","C","3","D","E","F"};
for (int i = 0; i < list.length; ) {
    int len = Integer.parseInt(list[i]);
    System.arraycopy(list, i + 1, list, i, len);
    list[i + len] = "-empty-";
    i += len + 1;
}
for (String value : list)
    System.out.println(value);

Output

A
B
-empty-
C
-empty-
D
E
F
-empty-

Hi Thank you for explaining the solution. I am new to Java. How do you improve problem solving skills or algorithm? Where can I practice this? — Aurora_Titanium, Sep 17 '15 at 06:17

Is it possible to use Integer.parse for String array of "mixed" types?

4 Answers4