leetCode 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 s, int numRows);

Example 1:

Input: s = "PAYPALISHIRING", numRows = 3
Output: "PAHNAPLSIIGYIR"

Example 2:

Input: s = "PAYPALISHIRING", numRows = 4
Output: "PINALSIGYAHRPI"
Explanation:

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

 1 //    方法一超过38.49%
 2     public String convert(String s, int numRows) {
 3         if (numRows <= 1 || s.length() <= 1) return s;
 4         char[][] res = new char[numRows][s.length() * (numRows - 1) / (2 * numRows - 2)+1];
 5         boolean direction = true;//行走方向，默认是向下走
 6         int row = 0, col = 0;//下一个填的位置
 7         for (int i = 0; i < s.length(); i++) {
 8             res[row][col] = s.charAt(i);
 9             if (row == numRows - 1) {//触底，改变行走方向
10                 direction = false;
11             }
12             if (i != 0 && row == 0) {//碰头，改变行走方向
13                 direction = true;
14             }
15             if (direction) {// 向下走
16                 ++row;
17             } else {// 向右上走
18                 --row;
19                 ++col;
20             }
21         }
22         s=new String("");
23         for (int i = 0; i < res.length; i++) {
24             for (int j = 0; j < res[0].length; j++) {
25                 if (res[i][j] != '')
26                     s += res[i][j];
27             }
28         }
29         return s;
30     }
31 
32 //  方法二：超过30.42%
33     String strOfi(String s, int begin, int step) {
34         String rs = new String("");
35 
36         while (begin < s.length()) {
37             rs += s.charAt(begin);
38             begin += step;
39         }
40         return rs;
41     }
42     public String convert(String s, int numRows) {
43         if (numRows <= 1) return s;
44         String rs = new String("");
45         //第一行
46         rs += strOfi(s, 0, 2 * numRows - 2);
47         //中间行
48         for (int i = 1; i < numRows-1; i++) {
49             String s1 = new String("");
50             String s2 = new String("");
51             s1=strOfi(s,i,2*numRows-2);
52             s2=strOfi(s,2*numRows-i-2,2*numRows-2);
53 
54             int j,k;
55             for (j = 0,k=0; j < s1.length() && k<s2.length(); j++,k++) {
56                 rs+=s1.charAt(j);
57                 rs+=s2.charAt(k);
58             }
59             if (j < s1.length()){
60                 rs+=s1.charAt(j);
61             }
62             if (k<s2.length()){
63                 rs+=s2.charAt(k);
64             }
65         }
66         //最后一行
67         rs += strOfi(s, numRows - 1, 2 * numRows - 2);
68         return rs;
69     }