LeetCode #6: Zigzag Conversion (C/C++).

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source: https://leetcode.com/problems/zigzag-conversion/
C/C++ Solution to LeetCode problem 6. Zigzag Conversion.

Problem

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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);

Examples

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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

Example 3:

Input: s = “A”, numRows = 1
Output: “A”

Constraints

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• 1 <= s.length <= 1000
• s consists of English letters (lower-case and upper-case), ',' and '.'.
• 1 <= numRows <= 1000

Solution

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The approach is simple, imagine a rectified sine wave, for example, if we want four rows:

“P A Y P A L I S H I R I N G” <- Original string
1 2 3 4 3 2 1 2 3 4 3 2 1 2 <- Row for each character
Distance for the next character for each row.
1: 6, 6, 6, 6, 6, 6….
2: 4, 2, 4, 2, 4, 2….
3: 2, 4, 2, 4, 2, 4….
4: 6, 6, 6, 6, 6, 6….

We can calculate the position of each letter in the original string, for the row we want.

• With this, then we just iterate from 0 to "size - 1" (the size of the string).
• For each position, calculate the position of the next character to append to the row we are currently forming.
• Once the calculated position for the next character is outside the size of the string, we finish with that row and we start forming the next one.

class Solution {
public:
  string convert(string s, int numRows) {
    if (numRows <= 1)
      return s;
      
    string result = ""s;

    int maxDistance = (numRows * 2) - 2;
    int maxRowDistance = maxDistance;
    int nextLetterDistance = maxDistance;
    int row = 0;
    int offset = 0;

    for (int i=0; i<s.size(); i+=1) {
      result += s[row + offset];
      offset += nextLetterDistance;

      nextLetterDistance = maxDistance - nextLetterDistance;
      if (nextLetterDistance == 0)
        nextLetterDistance = maxRowDistance;

      if (row + offset >= s.size()) {
        row += 1;
        offset = 0;
        maxRowDistance = (2 * (numRows - (row+1)));
        if (maxRowDistance == 0)
          maxRowDistance = maxDistance;
        nextLetterDistance = maxRowDistance;
      }
    }

    return result;
  }
};