Zigzag Conversion Leetcode Solution | Easy Approach

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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

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

Zigzag Conversion Solutions

Time: O(n)
Space: 

C++

class Solution {
 public:
  vector<int> twoSum(vector<int>& nums, int target) {
    unordered_map<int, int> numToIndex;

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 auto& row : rows)
      for (const char c : row)
        ans += c;

    return ans;
  }
};

Java

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

    return sb.toString();
  }
}

Python

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)

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