package com.fishercoder.solutions;

/**
 * 6. ZigZag Conversion
 *
 * 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".
 */

public class _6 {
	public static class Solution1 {
		public String convert(String s, int numRows) {
			StringBuilder[] sb = new StringBuilder[numRows];
			char[] c = s.toCharArray();
			int len = s.length();
			for (int i = 0; i < numRows; i++) {
				sb[i] = new StringBuilder();//this is an important step to initialize it
			}
			int i = 0;
			while (i < len) {
				for (int index = 0; index < numRows && i < len; index++) {
					sb[index].append(c[i++]);// vertically down
				}

				for (int index = numRows - 2; index >= 1 && i < len; index--) {
					/**Why it should start from numRows - 2? Think of the example when numRows = 3
					 the starting point of obliquely going up is 1, which is numRows-2.*/
					sb[index].append(c[i++]);// obliquely up
				}
			}

			for (i = 1; i < numRows; i++) {
				sb[0].append(sb[i]);
			}
			return sb[0].toString();
		}
	}

}

package com.fishercoder.solutions;

public class _6 {
	public static class Solution1 {
		public String convert(String s, int numRows) {
			StringBuilder[] sb = new StringBuilder[numRows];
			char[] c = s.toCharArray();
			int len = s.length();
			for (int i = 0; i < numRows; i++) {
				sb[i] = new StringBuilder();//this is an important step to initialize it
			}
			int i = 0;
			while (i < len) {
				for (int index = 0; index < numRows && i < len; index++) {
					sb[index].append(c[i++]);// vertically down
				}

				for (int index = numRows - 2; index >= 1 && i < len; index--) {
					/**Why it should start from numRows - 2? Think of the example when numRows = 3
					 the starting point of obliquely going up is 1, which is numRows-2.*/
					sb[index].append(c[i++]);// obliquely up
				}
			}

			for (i = 1; i < numRows; i++) {
				sb[0].append(sb[i]);
			}
			return sb[0].toString();
		}
	}

}

refactor 6