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：开始没有清晰的推导规律的思路，先要仔细观察，明确怎么去确定第i行包含S中哪些位置的元素

class Solution {
public:
    string convert(string s, int nRows)
    {
        if(nRows <= 1 || s.length() < 3 || s.length() <= nRows) return s;
        string s2;
        int zigSpan = nRows*2 - 2;
        for (int i = 0; i < nRows; i++)
        {
            for (int j = i; j < s.length(); j+=zigSpan)
            {
                s2.push_back(s[j]);  //先打印间隔相等为zigspan的垂直元素,第i行第一个元素为s[i],下一个s[i+zigspan]
                //第i行斜线第一个元素的位置nrow-(i+1)+nrow-1=2nrow-2-i,斜线相邻元素相隔zigspan=2nrow-2
                //第i行应该是打印，2nrow-2-i+0*zigspan，2nrow-2-i+1*zigspan,2nrow-2-i+2*zigspan,....,2nrow-2-i+n*zigspan
                //循环中,n=(j-i)*span,起始位置j-i=0
                if (i != 0 && i != nRows-1 && zigSpan+j-2*i<s.length())
                {
                    s2.push_back(s[zigSpan+j-2*i]);
                }
            }
        }
        return s2;
    }
};