0006. Zig Zag Conversion

# 6. ZigZag Conversion#

## 题目 #

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
``````

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`

## 题目大意 #

``````P   A   H   N
A P L S I I G
Y   I   R
``````

``````string convert(string s, int numRows);
``````

## 解题思路 #

• 这一题没有什么算法思想，考察的是对程序控制的能力。用 2 个变量保存方向，当垂直输出的行数达到了规定的目标行数以后，需要从下往上转折到第一行，循环中控制好方向ji

## 代码 #

``````package leetcode

func convert(s string, numRows int) string {
matrix, down, up := make([][]byte, numRows, numRows), 0, numRows-2
for i := 0; i != len(s); {
if down != numRows {
matrix[down] = append(matrix[down], byte(s[i]))
down++
i++
} else if up > 0 {
matrix[up] = append(matrix[up], byte(s[i]))
up--
i++
} else {
up = numRows - 2
down = 0
}
}
solution := make([]byte, 0, len(s))
for _, row := range matrix {
for _, item := range row {
solution = append(solution, item)
}
}
return string(solution)
}
``````

Apr 8, 2023