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

Example 3:

Input: s = "A", numRows = 1
Output: "A"

Constraints:

1 <= s.length <= 1000
s consists of English letters (lower-case and upper-case), ',' and '.'.
1 <= numRows <= 1000

---

z字型，新的字符串每行字符，每个字符对应左列开头字符以间隔 interval=2numRows-2 。每行每个周期会有2个字符不断以 interval 输出，第一行和最后一行比较特殊，每个周期只有1个。

class Solution {
    public String convert(String s, int numRows) {
        if(numRows==1) return s;
        int len=s.length();
        char[] c=new char[len];
        int interval=2*numRows-2;
        int index=0;
        int i1=0,i2;
        while(i1<len){
            c[index++]=s.charAt(i1);
            i1+=interval;
        }
        for(int i=1;i<numRows-1;++i){
            i1=i;
            i2=2*numRows-2-i1;
            while(true){
                if(i1>=len) break;
                c[index++]=s.charAt(i1);
                i1+=interval;
                if(i2>=len) break;
                c[index++]=s.charAt(i2);
                i2+=interval;
            }
        }
        i1=numRows-1;
        while(i1<len){
            c[index++]=s.charAt(i1);
            i1+=interval;
        }
        return String.valueOf(c);
    }
}