[LeetCode]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"

.

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：

1				9				17				25
2			8	10			16	18			24	26
3		7		11		15		19		23		27		…
4	6			12	14			20	22			28	30
5				13				21				29

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;
}
};