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”.
int main()
{
string s;
int nrow;
while (cin >> s >> nrow)
{
char imap[1000][1000]; //数组大小尝试出来的
memset(imap, 0, sizeof(imap));
int pos = 0;
int i = 0;
while (pos < s.size())
{
if (i % 2 == 0) //列为偶数时
{
for (int k = 0; k < nrow && s[pos]!='\0'; k++) //注意添加的时候,要判断s[pos]!='\0' ,如果等于的话,则也跳出了循环,因为pos此时等于s.size()了
{
imap[k][i] = s[pos++];
}
}
else //列为奇数时
{
if(s[pos] != '\0') //判断s[pos]!='\0'
{
if (nrow == 1) //nrow == 1另外考虑 ABCD 1 --> ABCD
imap[0][i] = s[pos++];
else if (nrow == 2) //nrow == 2另外考虑 ABCD 2 -->ACBD
for (int k = 0; k < nrow && s[pos] != '\0'; k++)
{
imap[k][i] = s[pos++];
}
else //nrow > 2, 从nrow - 2由下往上排
for (int k = nrow - 2; k >= 1 && s[pos] != '\0'; k--)
{
imap[k][i] = s[pos++];
}
}
}
i++;
}
s.clear();
for (int r = 0; r < nrow; r++)
for (int c = 0; c < i; c++)
{
if (imap[r][c] != 0)s.push_back(imap[r][c]);
}
cout << s << endl;
}
}