Zigzag Conversion Problem LeetCode (Print Concatenation of Zig-Zag String in 'n' Rows)

The string "PAYPALISHIRING" is written in a zigzag pattern on a given number of rows like this: (you may want to display this pattern in a fixed font for better legibility)

P   A   H   N
A P L S I I G
Y   I   R

And then read line by line: "PAHNAPLSIIGYIR"

Write the code that will take a string and make this conversion given a number of rows:

string convert(string text, int nRows);

convert("PAYPALISHIRING", 3) should return "PAHNAPLSIIGYIR".

This problem is popular in LeetCode, GeeksForGeeks, and StackOverFlow

SOLUTION:
Source code: Java

	import java.text.DateFormat;
	import java.text.SimpleDateFormat;
	import java.util.ArrayList;
	import java.util.Date;
	import java.util.HashMap;
	import java.util.List;
	import java.util.Map;
	/**
	* The string "PAYPALISHIRING" is written in a zigzag pattern on a given number of rows like this: (you may want to display this pattern in a fixed font for better legibility)
	P A H N
	A P L S I I G
	Y I R
	And then read line by line: "PAHNAPLSIIGYIR"
	Write the code that will take a string and make this conversion given a number of rows:
	string convert(string text, int nRows);
	convert("PAYPALISHIRING", 3) should return "PAHNAPLSIIGYIR".
	*/
	/**
	* @author rbaral
	*
	*/
	public class ZigzagConversion {
	public static void main(String[] args) {
	boolean testEnabled = true;
	if(testEnabled){
	performTest();
	}else{
	DateFormat format = new SimpleDateFormat("YYYY-M-d:H:m:s:S");
	System.out.println(format.format(new Date()));
	String s="PAYPALISHIRING";
	int rows=3;
	System.out.println("output is:"+convert(s, rows));
	System.out.println(format.format(new Date()));
	}
	}
	public static String convert(String s, int numRows) {
	if(numRows<2)
	return s;
	//lets make a list of char from the string array
	List<String> charList = new ArrayList<String>();
	for(int i=0;i<s.length();i++)
	charList.add(String.valueOf(s.charAt(i)));
	List<String> outputList = new ArrayList<String>();
	int rowIndex=0,nextElementIndex=0;
	for(int i=0;i<s.length();i++){
	//first add one element to each row
	if(rowIndex<numRows){
	//if this is the first time we are adding to this index
	if(outputList.size()<numRows)
	outputList.add(String.valueOf(s.charAt(i)));
	else
	outputList.set(rowIndex,outputList.get(rowIndex)+String.valueOf(s.charAt(i)));
	rowIndex++;
	nextElementIndex=0;
	}
	else{/*now if there are still elements, then append the the list from backwards, skipping the last and the first entries
	if the index has value 0, then the next element will be added to position 1
	find the position to add the remaining elements*/
	if(nextElementIndex==0)//this is the first time we are skipping characters
	nextElementIndex = numRows-1;//next element will be added to this row
	outputList.set(nextElementIndex-1,outputList.get(nextElementIndex-1)+String.valueOf(s.charAt(i)));
	if(nextElementIndex==1){//we reached the first row, so the next character should be inserted in second position
	rowIndex=1;
	}else
	nextElementIndex--;
	}
	}
	String output="";
	for(String item:outputList){
	output+=item;
	}
	return output;
	}
	public static void performTest(){
	/**
	* * Test cases:
	* abc,3->abc
	* abc,2->"acb"
	* PAYPALISHIRING,3->PAHNAPLSIIGYIR
	*/
	Map<String,String> testCases= new HashMap<String,String>();
	List<String> failedCases = new ArrayList<String>();
	String result="";
	testCases.put("abc,3", "abc");
	testCases.put("abc,2", "acb");
	testCases.put("PAYPALISHIRING,3", "PAHNAPLSIIGYIR");
	testCases.put("ABCD,2", "ACBD");
	testCases.put("PAYPA,2","PYAAP");
	testCases.put("ABCDE,4", "ABCED");
	testCases.put("ABCDEE,4", "ABECED");
	for(String testCase: testCases.keySet()){
	result = convert(testCase.split(",")[0], Integer.parseInt(testCase.split(",")[1]));
	if(!result.equals(testCases.get(testCase))){
	failedCases.add("Failed for:{"+testCase+"} got:"+result+" expected:"+testCases.get(testCase));
	}
	}
	System.out.println("TEST RESULT, Passed:"+(testCases.size()-failedCases.size())+" failed:"+failedCases.size());
	for(String failure:failedCases)
	System.err.println(failure);
	}
	}

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