LeetCode 6: Zigzag Conversion Solution

Master LeetCode problem 6 (Zigzag Conversion), a medium challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

6. Zigzag Conversion

Medium

String

Problem Explanation

Explanation:

To convert the given string into a zigzag pattern with a specified number of rows, we can simulate the zigzag pattern by iterating through the string and placing characters in the corresponding row. The approach involves keeping track of the current row for each character and then concatenating the characters row by row to form the final zigzag conversion.

We initialize an array of StringBuilder objects to represent each row.
We iterate through the characters of the input string and append each character to the corresponding row in the zigzag pattern.
We handle the direction changes at the beginning and end rows to zigzag back and forth.
Finally, we concatenate the rows to form the zigzag conversion string.

Time Complexity: O(n), where n is the length of the input string. Space Complexity: O(n), where n is the length of the input string.

Solution Code

class Solution {
    public String convert(String s, int numRows) {
        if (numRows == 1 || numRows >= s.length()) {
            return s;
        }
        
        StringBuilder[] rows = new StringBuilder[numRows];
        for (int i = 0; i < numRows; i++) {
            rows[i] = new StringBuilder();
        }
        
        int currentRow = 0;
        boolean goingDown = false;
        
        for (char c : s.toCharArray()) {
            rows[currentRow].append(c);
            
            if (currentRow == 0 || currentRow == numRows - 1) {
                goingDown = !goingDown;
            }
            
            currentRow += goingDown ? 1 : -1;
        }
        
        StringBuilder result = new StringBuilder();
        for (StringBuilder row : rows) {
            result.append(row);
        }
        
        return result.toString();
    }
}

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