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LeetCode Solutions

6. Zigzag Conversion

Time: $O(|\texttt{s}|)$

Space: $O(|\texttt{s}|)$

			


class Solution {
 public:
  string convert(string s, int numRows) {
    string ans;
    vector<vector<char>> rows(numRows);
    int k = 0;
    int direction = (numRows == 1) - 1;

    for (const char c : s) {
      rows[k].push_back(c);
      if (k == 0 || k == numRows - 1)
        direction *= -1;
      k += direction;
    }

    for (const vector<char>& row : rows)
      for (const char c : row)
        ans += c;

    return ans;
  }
};




			


class Solution {
  public String convert(String s, int numRows) {
    StringBuilder sb = new StringBuilder();
    List<Character>[] rows = new List[numRows];
    int k = 0;
    int direction = numRows == 1 ? 0 : -1;

    for (int i = 0; i < numRows; ++i)
      rows[i] = new ArrayList<>();

    for (final char c : s.toCharArray()) {
      rows[k].add(c);
      if (k == 0 || k == numRows - 1)
        direction *= -1;
      k += direction;
    }

    for (List<Character> row : rows)
      for (final char c : row)
        sb.append(c);

    return sb.toString();
  }
}




			


class Solution:
  def convert(self, s: str, numRows: int) -> str:
    rows = [''] * numRows
    k = 0
    direction = (numRows == 1) - 1

    for c in s:
      rows[k] += c
      if k == 0 or k == numRows - 1:
        direction *= -1
      k += direction

    return ''.join(rows)