## Zigzag Conversion

• Time:O(|\texttt{s}|)
• Space:O(|\texttt{s}|)

## C++

``````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()) {
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();
}
}
``````

## 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)
``````