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 s, int numRows);
Example 1:
Input: s = “PAYPALISHIRING”, numRows = 3
Output: “PAHNAPLSIIGYIR”
Example 2:
Input: s = “PAYPALISHIRING”, numRows = 4
Output: “PINALSIGYAHRPI”
Explanation:
P I N
A L S I G
Y A H R
P I
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 s, int numRows);
Example 1:
Input: s = “PAYPALISHIRING”, numRows = 3
Output: “PAHNAPLSIIGYIR”
Example 2:
Input: s = “PAYPALISHIRING”, numRows = 4
Output: “PINALSIGYAHRPI”
Explanation:
P I N
A L S I G
Y A H R
P I
题目大致意思：就是将一个字符串竖着以Z字型排列。
思路：其实可以用一个二维vector模拟，不过是竖着模拟。注意行越界的情况，还有当字符串是空的情况。
代码：
class Solution {
public:
string convert(string s, int numRows)
{
int len = s.length();
if(numRows==1) return s;
if(len==0) return "0";
vector<vector<char> > vec;
vec.resize(numRows);//r行
for (int k = 0; k < numRows; ++k) {
vec[k].resize(len);
}
for(int i=0;i<numRows;i++)
for(int j=0;j<vec[0].size();j++) vec[i][j] = '#';
cout<<vec[0].size()<<endl;
int row=0,col=0,cnt=0;
while(len)
{
if(col%(numRows-1)==0)
{
vec[row][col] = s[cnt];
row++;
}
else
{
vec[row][col] = s[cnt];
row--;
col++;
}
if(row==numRows)
{
row-=2;
col++;
}
--len;
++cnt;
}
cout<<col<<endl;
string ans;
for(int i=0;i<numRows;i++)
for(int j=0;j<=col;j++)
{
if(vec[i][j]!='#') ans.push_back(vec[i][j]);
}
return ans;
}
};