Leetcode:6. ZigZag Conversion

参考：http://blog.csdn.net/linhuanmars/article/details/21145039#reply，找规律

public String convert(String s, int nRows) {
    if(s == null || s.length()==0 || nRows <=0)
        return "";
    if(nRows == 1)
        return s;
    StringBuilder res = new StringBuilder();
    int size = 2*nRows-2;
    for(int i=0;i<nRows;i++)
    {
        for(int j=i;j<s.length();j+=size)
        {
            res.append(s.charAt(j));
            if(i!=0 && i!=nRows-1 && j+size-2*i<s.length())
                res.append(s.charAt(j+size-2*i));
        }                
    }
    return res.toString();
}