leetcode Question 125: ZigZag Conversion

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".

Analysis:

The idea is straightforward. Use a 2-D vector storing the zigzag sequence, for every element in the string, put it into proper position. Then output.
There are two major steps to put element into its position:
    (1).  Write # of nRows chars into the same column. like " | "
    (2).  Write # of nRows-2 chars into different column and rows in reverse diagonal order.   like " / ".

e.g.
Input: PAYPALISHIRING

2D vector:
0  1  2 3 4  5 6  7
-------------------
0 | P         I         N
1 | A     L S     I  G
2 | Y  A   H  R
3 | P         I

(updated 201309):

Note:  "ABCD" is    A  C
                               | / |
                              B  D

Code:

class Solution {
public:
    string convert(string s, int nRows) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function    
        vector<string> zz(nRows,"");
        if (nRows==1){return s;}
        
        int i=0; //for string 
        int r=0; //for zz vector
        bool zig = false; // if s[i] is in "/" or "|" direction
        while (i<s.size()){
            if (r<nRows && !zig){
                zz[r]+=s[i];
                r++;
                i++;
            }else{
                if (!zig){
                    zig=true;
                    r--;
                }else{
                    r--;
                    zz[r] = zz[r]+s[i];
                    if (r==0){zig=false;r++;}
                    i++;
                }
            }
        }
        string res; //get result;
        for (int i=0;i<zz.size();i++){
            for (int j=0;j<zz[i].size();j++){
                    res = res+ zz[i][j];
            }
        }
        return res;
    }
};

(old version)Code:

class Solution {
public:

    string printStr(vector<vector<char> > &t, int col,int nRow){
        string str="";
            for (int j=0;j<nRow;j++){
                for (int i =0; i<=col;i++){
                if (t[j][i]!='#'){
                    str = str+ t[j][i];
                }
            }
        }
        return str;
    }
    
    string convert(string s, int nRows) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function    
        if ((nRows<=0) || (s.size()==0)){return "";}
        if (nRows==1) {return s;}
        if (nRows>=s.size()){return s;}
        int k=s.size();
        vector<vector<char> > t(nRows,vector<char>(k,'#'));
        int len = s.size();
        int col = 0;
        int row = 0;
        int count = 0;
        while(count < len){
            if (row<nRows){
                t[row][col]=s[count];
                row++;
                count++;
            }else{
                col++;
                for (int j = nRows-2; j>=1;j--){
                    t[j][col]=s[count];
                    count ++;
                    if (count>=len){ return  printStr(t,col,nRows);}
                    col++;
                }
                row = 0;
            }
        }
        return printStr(t,col,nRows);
    }
};