# Zigzag Conversion – Leetcode Solution

In this post, we are going to solve the 6. Zigzag Conversion problem of Leetcode. This problem 6. Zigzag Conversion is a Leetcode medium level problem. Let’s see code, 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);`

### 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`

Now, let’s see the code of 6. Zigzag Conversion – Leetcode Solution.

# Zigzag Conversion – Leetcode Solution

### 6. Zigzag Conversion – Solution in Java

```class Solution {
public String convert(String s, int nRows) {
char[] c = s.toCharArray();
int len = c.length;
StringBuffer[] sb = new StringBuffer[nRows];
for (int i = 0; i < sb.length; i++) sb[i] = new StringBuffer();

int i = 0;
while (i < len) {
for (int idx = 0; idx < nRows && i < len; idx++) // vertically down
sb[idx].append(c[i++]);
for (int idx = nRows-2; idx >= 1 && i < len; idx--) // obliquely up
sb[idx].append(c[i++]);
}
for (int idx = 1; idx < sb.length; idx++)
sb[0].append(sb[idx]);
return sb[0].toString();
}
}```

### 6. Zigzag Conversion – Solution in C++

```class Solution {
public:
string convert(string s, int numRows) {
vector<string> vs(numRows, "");
int n = s.length(), i = 0;
while (i < n) {
for (int j = 0; j < numRows && i < n; j++)
vs[j].push_back(s[i++]);
for (int j = numRows - 2; j >= 1 && i < n; j--)
vs[j].push_back(s[i++]);
}
string zigzag;
for (string v : vs) zigzag += v;
return zigzag;
}
};```

### 6. Zigzag Conversion– Solution in Python

```class Solution(object):
def convert(self, s, numRows):
"""
:type s: str
:type numRows: int
:rtype: str
"""
if numRows == 1 or numRows >= len(s):
return s

L = [''] * numRows
index, step = 0, 1

for x in s:
L[index] += x
if index == 0:
step = 1
elif index == numRows -1:
step = -1
index += step

return ''.join(L)```

Note: This problem 6. Zigzag Conversion is generated by Leetcode but the solution is provided by CodingBroz. This tutorial is only for Educational and Learning purpose.