6. ZigZag Conversion

https://leetcode.com/problems/zigzag-conversion/description/
The gist of the title: Given a string and n, it is required to output according to the "zi" glyph of n lines. Problem-
solving ideas: as shown below:

/*n=numRows
Δ=2n-2    0                           2n-2                           4n-4
Δ=        1                     2n-3  2n-1                    4n-5   4n-3
Δ=        2               2n-4        2n                 4n-6         .
Δ=        .             .               .               .             .
Δ=        .       n+1                 .           3n-1                .
Δ=        n-2   n                     3n-4    3n-2                   5n-6
Δ=2n-2    n-1                         3n-3                           5n-5
*/

The difference between the two columns is 2n-2 (n is the row number), except for the 0th row and the n-1th row, the rest have a "zigzag" shape, such as 2n-3 in the 1st row, but between the two columns The difference remains unchanged and is still 2n-2. The formula for the intermediate quantity is:

  +2（n-1-1）       +2*1
1-----------2n-3----------2n

That is, the first step+2(n-1-i), the second step+2*i, i is the row number.
That's it.
In addition, special cases must be considered: numRows=1, which will enter an infinite loop, so special treatment is required.

Code:

class Solution {
public:
    string convert(string s, int numRows) {
        if (numRows == 1) return s;  //特殊处理
        int len = s.length();
        string result = "";
        for (int i = 0; i < numRows; i++) {
            int step1 = 2*(numRows-1-i);
            int step2 = 2*i;
            int pos = i;
            result += s[pos];
            while (1) {
                pos += step1;
                if (pos >= len) break;
                if (step1) result += s[pos];  //防止重复输出（n=0和n=numRows-1）
                pos += step2;
                if (pos >= len) break;
                if (step2) result += s[pos];  //防止重复输出
            }
        }
        return result;
    }
};