【Leetcode】Zigzag Conversion

【题目】

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 text, int nRows);

convert("PAYPALISHIRING", 3) should return "PAHNAPLSIIGYIR".

【思路】

The problem statement itself is unclear for many. Especially for 2-row case. "ABCD", 2 --> "ACBD". The confusion most likely is from the character placement. I would like to extend it a little bit to make ZigZag easy understood.

The example can be written as follow:

1. P.......A........H.......N
2. ..A..P....L..S....I...I....G
3. ....Y.........I........R

Therefore, <ABCD, 2> can be arranged as:

1. A....C
2. ...B....D

【代码】

     public String convert(String s, int numRows) {
        String[]lines=new String[numRows];
        
        for (int i=0;i<numRows;i++){
            lines[i]="";
        }
        char[] a=s.toCharArray();
        int dir=-1;
        int row=0;
        for (int i=0;i<a.length;i++){
            lines[row]+=a[i];
            if(row==0 || row==numRows-1)
                dir*=-1;
            row+=dir;
            row%=numRows;
        }

        String res="";
        for (String line : lines)
            res+=line;
        return res;

    }

string convert(string s, int nRows) {

    if (nRows <= 1)
        return s;

    const int len = (int)s.length();
    string *str = new string[nRows];

    int row = 0, step = 1;
    for (int i = 0; i < len; ++i)
    {
        str[row].push_back(s[i]);

        if (row == 0)
            step = 1;
        else if (row == nRows - 1)
            step = -1;

        row += step;
    }

    s.clear();
    for (int j = 0; j < nRows; ++j)
    {
        s.append(str[j]);
    }

    delete[] str;
    return s;
}