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 text, int nRows);

convert("PAYPALISHIRING", 3) should return "PAHNAPLSIIGYIR". 

思考过程:以题目所举例子为例，第一个字符放在第一行，第二个放在第二行，...。这一个过程是以4为一个循环，那么能否对字符串中的任意个字符预先判断出其所在的行数呢，通过小例子的演算，发现，字符的下标对4取余，即可知道其所在的行数。据此实现程序。

char* convert(char* s, int numRows) {
    if(!strlen(s)) return s;
    
    if(numRows==1) return s;
    
    int len = strlen(s);
    char *s2 = malloc(sizeof(char) * (len+1));
    memset(s2, 0, sizeof(s2));

    int count = 0;
    
    int loop = numRows + (numRows -2);
    for(int i= 0; i<numRows; i++) {
        for(int j = 0; j<len; j++){
            if(j%loop==i || j%loop==loop-i){
                s2[count]=s[j];
                count++;
            }
        }
    }
    
    s2[count]='';
    
    return s2;
}