Question

Formatted question description: https://leetcode.ca/all/6.html

 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

Algorithm

Law: The first and last lines are added in the order of 2n-2. The position of the word diagonally across the line is the current column j+(2n-2)-2i (i is the index of the row).

Code

Java

public class ZigZag_Conversion {

    public class Solution {
        public String convert(String s, int nRows) {
            if (s == null || s.length() <= nRows || nRows == 1) {
                return s;
            }
            StringBuilder sb = new StringBuilder();
            for (int i = 0; i < nRows; i++) {
                if (i == 0 || i == nRows - 1) {
                    int index = i;
                    while (index < s.length()) {
                        sb.append(s.charAt(index));
                        index += 2 * (nRows - 1);
                    }
                } else {
                    int index = i;
                    while (index < s.length()) {
                        sb.append(s.charAt(index));
                        if (index + 2 * nRows - 2 * i - 2 < s.length()) {
                            sb.append(s.charAt(index + 2 * nRows - 2 * i - 2));
                        }
                        index += 2 * (nRows - 1);
                    }
                }
            }
            return sb.toString();
        }
    }
}

// OJ: https://leetcode.com/problems/zigzag-conversion/
// Time: O(N)
// Space: O(1)
class Solution {
public:
    string convert(string s, int numRows) {
        if (numRows == 1) return s;
        int N = s.size(), d = (numRows - 1) * 2;
        string ans;
        for (int i = 0; i < numRows; ++i) {
            int w = 2 * i; 
            for (int j = i; j < N;) {
                ans += s[j];
                w = d - w;
                if (!w) w = d;
                j += w;
            }
        }
        return ans;
    }
};

class Solution(object):
  def convert(self, s, numRows):
    """
    :type s: str
    :type numRows: int
    :rtype: str
    """
    if numRows <= 1:
      return s
    n = len(s)
    ans = []
    step = 2 * numRows - 2
    for i in range(numRows):
      one = i
      two = -i
      while one < n or two < n:
        if 0 <= two < n and one != two and i != numRows - 1:
          ans.append(s[two])
        if one < n:
          ans.append(s[one])
        one += step
        two += step
    return "".join(ans)

6 - ZigZag Conversion

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