[Leetcode 42] 6 ZigZag Conversion

Problem:

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".

Analysis:

First of all, you should know what's the meaning of zigzag. To understand more, try some more examples.

If n = 2, then "PAYPALISHIRING" is arranged into:

P Y A I H R N

A P L S I I G

and converted into "PYAIHRNAPLSIIG"

If n = 4, then "PAYPALISHIRING" is arranged into:

P         I         N

A     L  S     I  G

Y  A     H  R

P        I 

If n = 5, then  "PAYPALISHIRING" is arranged into:

P            H

A        S  I         G

Y      I     R    N

P  L         I

A

Now try to find the relationship of characters in the same row. We may find that, for elements in 

n = 2, two characters in the same row's index differs 2

n = 3, first line's characters' index differs 4; second line's characters' index differs 2 2; third line's characters' index differs 4;

n = 4, first line's characters' index differs 6; second line's characters' index differs 4 2; third line's characters' index differs 2 4; fourth line difffers 6;

n=5, first line differ 8; second line differ 6 2; third line differ 4 4; fourth line differ 2 6; fifth line differ 0 8;

It's clear that there is a relationship. If think carefully, you may find the relationship is as follows:

fp = 2*(n-1), sp=2*i; Then the next element is draw from current index plus fp/sp staggered.

So the solution is as follows.

Code:

class Solution {
public:
    string convert(string s, int nRows) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function    
        if (nRows == 1) return s;
        
        char res[s.length()];
        int idx = 0;
        
        for (int i=0; i<nRows; i++) {
            int fp = 2*(nRows - i - 1);
            int sp = 2*i;
            
            bool flag = true;
            for (int j=i; j<s.length(); ) {
                if (flag) {
                    if (fp != 0) 
                        res[idx++] = s[j];
                    j += fp;
                }
                else {
                    if (sp != 0)
                        res[idx++] = s[j];
                        
                    j += sp;
                }
                
                flag = !flag;
            }
        }
        res[idx] = '\0';
        
        string r(res);
        return r;
    }
};