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

翻译
将字符串 “PAYPALISHIRING” 以Z字形排列成给定的行数：
P A H N
A P L S I I G
Y I R
之后从左往右，逐行读取字符：“PAHNAPLSIIGYIR”
实现一个将字符串进行指定行数变换的函数:
string convert(string s, int numRows);

分析
首位两行间隔固定，中间行间隔线性变化，分类讨论即可。

c++实现

class Solution {
public:
    string convert(string s, int numRows) {
        if(numRows == 1)return s;
        int len = s.size(), k = 0, interval = (numRows<<1)-2;
        string res(len, ' ');
        for(int j = 0; j < len ; j += interval)//处理第一行
            res[k++] = s[j];
        for(int i = 1; i < numRows-1; i++)//处理中间行
        {
            int inter = (i<<1);
            for(int j = i; j < len; j += inter)
            {
                res[k++] = s[j];
                inter = interval - inter;
            }
        }
        for(int j = numRows-1; j < len ; j += interval)//处理最后一行
            res[k++] = s[j];
        return res;
    }
};