In this post, we are going to solve the Zigzag Conversion Leetcode Solution problem of Leetcode. This Leetcode problem is done in many programming languages like C++, Java, and Python.
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 <= 1000sconsists of English letters (lower-case and upper-case),','and'.'.1 <= numRows <= 1000
Now, let’s see the leetcode solution of Zigzag Conversion Leetcode Solution.
Zigzag Conversion Leetcode 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)Zigzag Conversion Leetcode Solution in CPP
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;
}
};Zigzag Conversion Leetcode 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();
}
}Note: This problem Zigzag Conversion is generated by Leetcode but the solution is provided by Chase2learn This tutorial is only for Educational and Learning purposes.
