6. ZigZag Conversion

https://leetcode.com/problems/zigzag-conversion/

C++

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

        string result;
        result.reserve(s.length());
        int size = 2 * numRows - 2;

        for (int i = 0; i < numRows; i++) {
            for (int j = i; j < s.length(); j += size) {
                result.push_back(s[j]);
                if (i > 0 && i < numRows - 1 && j + size - 2 * i < s.length()) {
                    result.push_back(s[j + size - 2 * i]);
                }
            }
        }

        return result;
    }
};

Go

import "strings"

func convert(s string, numRows int) string {
  if numRows <= 1 {
    return s
  }

  result := []string{}
  size := 2*numRows - 2

  for i := 0; i < numRows; i++ {
    for j := i; j < len(s); j += size {
      result = append(result, string(s[j]))
      if i > 0 && i < numRows-1 && j+size-2*i < len(s) {
        result = append(result, string(s[j+size-2*i]))
      }
    }
  }

  return strings.Join(result, "")
}

JavaScript

/**
 * @param {string} s
 * @param {number} numRows
 * @return {string}
 */
var convert = function(s, numRows) {
  if (numRows <= 1) {
    return s
  }

  var result = []
  var size = 2 * numRows - 2

  for (let i = 0; i < numRows; i++) {
    for (let j = i; j < s.length; j += size) {
      result.push(s[j])
      if (i > 0 && i < numRows - 1 && j + size - 2 * i < s.length) {
        result.push(s[j + size - 2 * i])
      }
    }
  }

  return result.join('')
}

Python

class Solution(object):

    def convert(self, s, numRows):
        """
        :type s: str
        :type numRows: int
        :rtype: str
        """
        if numRows <= 1:
            return s
        result = []
        size = 2 * numRows - 2
        for i in range(numRows):
            for j in range(i, len(s), size):
                result.append(s[j])
                if i > 0 and i < numRows - 1 and j + size - 2 * i < len(s):
                    result.append(s[j + size - 2 * i])
        return ''.join(result)