Leetcode Q6：ZigZag Conversion

题目6：

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"

.

/* 获取字符串长度 */
int GetLen(char* s)
{
int len = 0;
while (*(s++) != '\0')
{
len++;
}
return len;
}

char* convert(char* s, int numRows)
{
int i = 0;
int j = 0;
int p = 0;
int str_len = 0;
int sub_len = 0;
int x = 0;
int y = 0;
int row = 0;
int col = 0;
char** arr = NULL;

str_len = GetLen(s);

if (str_len == 0 || str_len == 1 || numRows == 1)
{
return s;
}

/* 计算长度  */
sub_len = (numRows - 1) * (str_len / (2 * numRows - 2) + 1);

arr = (char**)malloc(sizeof(char*)*numRows);
for (i = 0; i < numRows; i++)
{
*(arr+i) = (char*)malloc(sub_len);
for (j = 0; j < sub_len; j++)
{
*(*(arr+i) + j) = 0;
}
}
for (i = 0; i < str_len; i++)
{
x = i % (2 * numRows - 2);
y = i / (2 * numRows - 2);

/* 计算行号和列号 */
if (x <= numRows - 1)
{
row = x;
col = y * (numRows - 1);
}
else
{
row = 2 * numRows - 2 - x;
col = i % (numRows - 1) + (numRows - 1) * y;
}

*(*(arr + row) + col) = s[i];
}

/* 输出数组元素 */
for (i = 0; i < numRows; i++)
{
for (j = 0; j < sub_len; j++)
{
if ( *(*(arr + i) + j) != '\0')
{
s[p] = *(*(arr + i) + j);
p++;
}
}
}
return s;
}