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

medium

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

## Problem

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

### 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

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

## Solution

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.
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 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; } };