ZigZag Conversion

[ioleon13]

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".

[ioleon13]

思路：使用二维字符数组保存zigzag转换出来的字符串，其他位置使用０填充。通过nRows得出一次完整zigzag需要的元素个数nZigZagLen = 2 * nRows - 2，所以将字符串转换为zigzag字符串需要的循环次数nZigZagCount = string.len/nZigZagLen

[ioleon13]

每一次zigzag循环,矩阵中斜线部分字符的填充:
for(int i = 0; i < nRows-2; ++i) {
matrix[nRows-2-i][nZigZagPos*(nRows-1)+1+i] = s[nPos]; //nZigZagPos是上一层nZigZagCount循环的下标
nPos++;
}