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

Write the code that will take a string and make this conversion given a number of rows:

string convert(string text, int nRows);

class Solution {
public:
    string convert(string s, int nRows) {
    if(s.size() <= 1 || nRows == 1)
    {
        return s;
    }
    
    int totalCol = (s.size() + 2 * nRows - 1)/(2 * nRows - 2) * (nRows - 1);
	char **rct = new char* [nRows];
	//std::vector< std::vector<char> > rct;
	for (int i = 0; i < nRows; i++)
	{
		rct[i] = new char[totalCol];
		for (int j = 0; j < totalCol; j++)
		{
			rct[i][j] = ‘ ‘;
		}
	}
	
	//char rct[10][10];
	int col = 0;
	for (int i = 0; i < s.size(); i++)
	{
		/*int col = i / (2 * nRows - 2);*/
		if (i % (2 * nRows - 2) == 0 )
		{
			int R = 0;
			while(i < s.size() && R < nRows)
			{
				//第col列先赋值
				rct[R][col] = s[i];
				R++;
				i = i++;
				//对第col++列赋值
				if (R == nRows - 1)
				{
					while(R > 0 && i < s.size())
					{	
						rct[R][col] = s[i];
						i = i++;
						R--;
						col++;
					}
				}
				//R又回到0,跳出
				if (R == 0)
				{
					i--;
					break;
				}
			}
		}
	}
	std::string ans;
	for (int i = 0; i < nRows; i++)
	{
		for (int j = 0; j < totalCol; j++)
		{
			if (rct[i][j] != ‘ ‘)
			{
				ans += rct[i][j];
			}
		}
	}
	for (int i = 0; i < nRows; i++)
	{
		delete[] rct[i];
		rct[i] = NULL;
	}
	delete[] rct;
	rct = NULL;
	return ans;
        
    }
};

网上看到的简洁代码：

题意的 "z" 字形指一列nRows个，然后斜着往右上一格一个回到第一行，然后再一列nRows个。比如nRows=5，如下：

每行字母在原字符串中的间隔是有规律的，虽然两层for循环，但是s中每个字母只访问了一次，O(n)。

class Solution {
public:
    string convert(string s, int nRows) {
        if(nRows == 1) return s;
        string ans;
        int a = (nRows << 1) - 2, b = 0;
        for(int i = 0; i < nRows; i ++, a -= 2, b += 2)
        {
            bool flag = false;
            for(int j = i; j < s.length(); 
                    j += flag ? (b ? b : a) : (a ? a : b), flag ^= 1)
                ans += s[j];
        }
        return ans;
    }
};