6. ZigZag Conversion

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

string convert(string s, int numRows) { // time: O(n); space: O(n)
    if (s.empty()) return "";
    int n = s.length(), i = 0;
    vector<string> tmp(numRows, "");
    while (i < n) {
        // Vertical
        for (int j = 0; j < numRows && i < n; ++j) {
            tmp[j].push_back(s[i++]);
        }
        // Oblique
        for (int j = numRows - 2; j >= 1 && i < n; --j) {
            tmp[j].push_back(s[i++]);
        }
    }
    string res;
    for (const string& str : tmp) res += str;
    return res;
}

string convert(string s, int numRows) { // time: O(n); space: O(n)
    if (s.empty() || numRows <= 1) return s;
    int n = s.length(), size = 2 * numRows - 2;
    string res;
    for (int i = 0; i < numRows; ++i) {
        for (int j = i; j < n; j += size) {
            res += s[j];
            int pos = j + size - 2 * i;
            if (i != 0 && i != numRows - 1 && pos < n) res += s[pos];
        }
    }
    return res;
}