6.Zigzag Conversion
===
###### tags: `Medium`,`String`
[6. Zigzag Conversion](https://leetcode.com/problems/zigzag-conversion/)
### 題目描述
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);
```
### 範例
**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
```
**Example 3:**
```
Input: s = "A", numRows = 1
Output: "A"
```
**Constraints**:
* 1 <= `s.length` <= 1000
* `s` consists of English letters (lower-case and upper-case), `','` and `'.'`.
* 1 <= `numRows` <= 1000
### 解答
#### Python
```python=
class Solution:
def convert(self, s: str, numRows: int) -> str:
if numRows == 1:
return s
buf = []
for _ in range(numRows):
buf.append( [ 0 for _ in range(len(s))] )
row,col = 0, 0
direction = 'down'
for c in s:
#print(row,col,direction)
buf[row][col] = c
if row == numRows-1:
direction = 'up'
row, col = row-1, col+1
elif row == 0:
direction = 'down'
row, col = row+1, col
else:
if direction == 'down':
row, col = row+1, col
else:
row, col = row-1, col+1
result = ''
for i in range(numRows):
for j in range(len(s)):
#print(buf[i][j], end='')
if buf[i][j] != 0:
result += buf[i][j]
#print('')
return result
```
> [name=玉山]
#### C++
```cpp=
class Solution {
public:
string convert(string s, int numRows) {
if (numRows == 1) return s;
string ans;
int step = (numRows - 1) * 2;
for (int i = 0; i < numRows; i++) {
for (int j = 0; j < s.size(); j += step) {
int left = j + i;
int right = j + step - i;
if (left < s.size()) ans += s[left];
if (right < s.size() && left < right && right < j + step) ans += s[right];
}
}
return ans;
}
};
```
> [name=Yen-Chi Chen][time=Fri, Feb 3, 2023]
### Reference
[回到題目列表](https://hackmd.io/@Marsgoat/leetcode_every_day)