6. ZigZag Conversion

    /*
     * 6. ZigZag Conversion
     * 2016-8-26 by Mingyang
     * 觉得这个题目面试的必要性不大，有规律，第一行和最后一行每两个相差2n-2
     * 中间的每行除了2n-2有外，中间的一个也有2 * (n - 1 - i)
     */
       public String convert(String s, int numRows) {
            if (numRows == 1)
                return s;
            StringBuilder sb = new StringBuilder();
            // step
            int step = 2 * numRows - 2;
            for (int i = 0; i < numRows; i++) {
                //first & last rows
                if (i == 0 || i == numRows - 1) {
                    for (int j = i; j < s.length(); j = j + step) {
                        sb.append(s.charAt(j));
                    }
                //middle rows
                } else {
                    int j = i;
                    boolean flag = true;
                    int step1 = 2 * (numRows - 1 - i);
                    int step2 = step - step1;

                    while (j < s.length()) {
                        sb.append(s.charAt(j));
                        if (flag)
                            j = j + step1;
                        else
                            j = j + step2;
                        flag = !flag;
                    }
                }
            }
            return sb.toString();
        }