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string convert(string text, int nRows);

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Solution

Two ways to work out this problem.

Solution 1. Insert string s vertically into a 2d array, char by char. After this is finished, make the result string by reading the 2d array horizontally. See code 1 below.

Solution 2. Calculate and find out (in sequence) which char from s should be inserted into the result list. Then build the result list directly. See code 2 below.

My code

One. Fill in the whole string, then parse the result.

public class Solution {
    public String convert(String s, int nRows) {
        if (s == null || s.length() == 0 || nRows < 1) {
            return "";
        } else if (nRows == 1) {
            return s;
        }
        int len = s.length();
        int grpSize = nRows * 2 - 2;
        int numGrp = (len - 1) / grpSize + 1;
        String[][] array = new String[nRows][numGrp * 2];
        // fill in string s into array, one group after another
        int p = 0;
        // for each group
        for (int i = 0; i < numGrp; i++) {
            // for each vertical index (left col), fill in one letter from s
            for (int j = 0; j < nRows; j++) {
                // if s is used up, break
                if (p == len) {
                    break;
                }
                array[j][i * 2] = "" + s.charAt(p);
                p++;
            }
            // for each vertical index (right col), fill in one letter from s
            for (int j = nRows - 2; j > 0; j--) {
                // if s is used up, break
                if (p == len) {
                    break;
                }
                array[j][i * 2 + 1] = "" + s.charAt(p);
                p++;
            }
        }
        // now convert array[][] into final string and return answer
        StringBuilder sb = new StringBuilder();
        for (int i = 0; i < array.length; i++) {
            for (int j = 0; j < array[0].length; j++) {
                if (array[i][j] == null || array[i][j].length() == 0) {
                    continue;
                } else {
                    sb.append(array[i][j]);
                }
            }
        }
        return sb.toString();
    }
}

Two. Pick the correct char and form the result string, and fill it in the result string.

public class Solution {
    public String convert(String s, int nRows) {
        if (nRows <= 1)
            return s;
        int eachPattern = 2 * nRows - 2;
        int numPatterns = (s.length() - 1) / eachPattern + 1;
        StringBuilder ans = new StringBuilder();
        for (int j = 0; j < nRows; j++) {
            for (int i = 0; i < numPatterns; i++) {
                ans.append(find(s, eachPattern, i, j));
                if (j != 0 && j != nRows - 1)
                    ans.append(find(s, eachPattern, i, 2 * (nRows - 1) - j));
            }
        }
        return ans.toString();
    }

    private String find(String s, int eachPattern, int i, int j) {
        // find (j)th element from (i)th pattern
        int temp = eachPattern * i + j;
        if (temp < s.length())
            return s.substring(temp, temp + 1);
        return "";
    }
}