[leetcode] ZigZag Conversion

From :https://leetcode.com/problems/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".

Solution 1:

class Solution {
public:
    string convert(string s, int numRows) {
        if(numRows==1) return s;
		string result;
        int off = 2*numRows-2, index, len=s.length(), po=0;
        for(int i=0; i<numRows; i++) {
            int index = i;
            if(index==0 || index==numRows-1)  {
				while(index<len) {
					result += s[index]; 
					index += off;
				}
            } else {
                int midOff = numRows-i-1, preIndex;
                while(index<len) {
					result += s[index]; 
                    preIndex = index;
                    index = 2*(index+midOff)-index;
                    if(index<len) {
						result += s[index]; 
                    }
                    index = preIndex + off;
                }
            }
            
        }
        return result;
    }
};

class Solution {
public:
    string convert(string s, int rows) {
        if(rows == 1) {
            return s;
        }
        string ans;
        int len = s.size(), step = rows + rows - 2;
        for(int row=0; row<rows; ++row) {
            if(0 == row || row+1 == rows) {
                for(int i=row; i<len; i+=step) {
                    ans += s[i];
                }
            } else {
                int off = rows - row - 1;
                for(int i=row; i<len; i+=step) {
                    ans += s[i];
                    if(i+off+off < len) {
                        ans += s[i+off+off];
                    }
                }
            }
        }
        return ans;
    }
};

solution 2:

class Solution {
public:
    string convert(string s, int nRows) {
        // IMPORTANT: Please reset any member data you declared, as
        // the same Solution instance will be reused for each test case.  
        
        if (nRows <= 1 || s.length() == 0)
           return s;

        string res = "";
        int len = s.length();
        for (int i = 0; i < len && i < nRows; ++i)
        {
			int indx = i;
			res += s[indx];

            for (int k = 1; indx < len; ++k)
            {
                //第一行或最后一行，使用公式1：
                if (i == 0 || i == nRows - 1)
                {
                    indx += 2 * nRows - 2;
                }
                //中间行，判断奇偶，使用公式2或3
                else
                {
                    if (k & 0x1)  //奇数位
						indx += 2 * (nRows - 1 - i);
                    else indx += 2 * i;
                }

                //判断indx合法性
                if (indx < len)
                {
                    res += s[indx];
                }	
            }
        }
        return res;
    }
};