【刷题笔记】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

将一个给定字符串 s 根据给定的行数 numRows ，以从上往下、从左到右进行 Z 字形排列。

比如输入字符串为 "PAYPALISHIRING" 行数为 3 时，排列如下：

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

之后，你的输出需要从左往右逐行读取，产生出一个新的字符串，比如："PAHNAPLSIIGYIR"。

请你实现这个将字符串进行指定行数变换的函数：

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)
}