ZigZag Conversion @LeetCode

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".

	package leetcode.string;
	/**
	* Solution: Math 题，找index 规律，遍历, 根据nRow和ZigZag长度来给append每个元素, 所有节点访问一遍，Time O(n) Space O(1)
	* ZigZig size = 2*n-1. 最后一个节点会重复计算，所以我们用2*n - 2.
	* 满的列 j =i, index 为 j, j+size, j+2size....
	* 不满的 j-2*i+size.
	*
	* Reference:
	* http://fisherlei.blogspot.com/2013/01/leetcode-zigzag-conversion.html
	* http://codeganker.blogspot.com/2014/02/zigzag-conversion-leetcode.html
	* @author jeffwan
	* @date Apr 15, 2014
	*/
	public class ZigZagConversion {
	public String convert(String s, int nRows) {
	if (s == null || s.length() == 0 || nRows <= 0) {
	return "";
	}
	if (nRows == 1) {
	return s;
	}
	StringBuilder result = new StringBuilder();
	int size = 2 * nRows - 2; // ZigZag Length (without last duplicate node)
	for (int i = 0; i < nRows; i++) {
	for (int j = i; j < s.length(); j += size) {
	result.append(s.charAt(j));
	if (i != 0 && i != nRows - 1 && j + size - 2 * i < s.length()) {
	result.append(s.charAt(j + size - 2 * i));
	}
	}
	}
	return result.toString();
	}
	}

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