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leetcode 6. zigzag conversion

admin 11月 12, 2020 0

6. ZigZag Conversion

Difficulty: Medium

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)

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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:

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string convert(string s, int numRows);

Example 1:

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**Input:** s = "PAYPALISHIRING", numRows = 3
**Output:** "PAHNAPLSIIGYIR"

Example 2:

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**Input:** s = "PAYPALISHIRING", numRows = 4
**Output:** "PINALSIGYAHRPI"
**Explanation:**

P     I    N
A   L S  I G
Y A   H R
P     I

Solution

Language: Java

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class  {
    public String convert(String s, int numRows) {
        if (s == null || s.length() == 0 || numRows == 1) {
            return s;
        }
        int l = s.length();
        StringBuilder sb = new StringBuilder();
        int a = (numRows - 1) * 2;
        for (int i = 0; i < numRows; i++) {
            int b = a - 2 * i;
            for (int j = i; j < l; j += a) {
                sb.append(s.charAt(j));
                if (b != 0 && b != a && j + b < l) {
                    sb.append(s.charAt(j + b));
                }
            }
        }
        return sb.toString();
    }
}
 
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class  {
    public String convert(String s, int numRows) {
        if(s == null || s.length() == 0 || numRows <= 1) {
            return s;
        }
        StringBuilder[] builders = new StringBuilder[numRows];
        int cur = 0;
        int direction = 1;
        for (int i = 0; i < builders.length; i++) {
            builders[i] = new StringBuilder();
        }
        for (int i = 0; i < s.length(); i++) {
            builders[cur].append(s.charAt(i));
            cur += direction;
            if (cur == numRows) {
                cur -= 2;
                direction = -1;
            } else if (cur == -1) {
                cur += 2;
                direction = 1;
            }
        }
        for (int i = 1; i < builders.length; i++) {
            builders[0].append(builders[i]);
        }
        return builders[0].toString();
    }
}
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