golang 6 Z字形变换

Posted 2021-05-24

package main

import (
	"fmt"
	"strings"
)

func main() {
	s := "LEETCODEISHIRING"
	fmt.Println(convert(s, 4))
}
func convert(s string, numRows int) string {
	if numRows == 1 {
		return s
	}
	rows := make([]strings.Builder, 0)
	for i := 0; i < numRows; i++ {
		rows = append(rows, strings.Builder{})
	}
	curRow := 0
	goingDown := false

	for _, c := range s {
		rows[curRow].WriteByte(byte(c))
		if curRow == 0 || curRow == numRows-1 {
			goingDown = !goingDown
		}
		if goingDown {
			curRow++
		} else {
			curRow--
		}
	}
	ret := strings.Builder{}
	for _, v := range rows {
		ret.WriteString(v.String())
	}
	return ret.String()
}

Z字形变换

将字符串 "PAYPALISHIRING" 以Z字形排列成给定的行数：

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

之后从左往右，逐行读取字符："PAHNAPLSIIGYIR"

实现一个将字符串进行指定行数变换的函数:

string convert(string s, int numRows);

示例 1:

输入: s = "PAYPALISHIRING", numRows = 3
输出: "PAHNAPLSIIGYIR"

示例 2:

输入: s = "PAYPALISHIRING", numRows = 4
输出: "PINALSIGYAHRPI"
解释:

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

 1 class Solution(object):
 2     def convert(self,s,numRows):
 3         ‘‘‘
 4 
 5         :type s:str
 6         :type numRows:int
 7         :rtype:str
 8         ‘‘‘
 9         if numRows==1 or numRows >= len(s):
10             return s
11         L=[‘‘]*numRows #建立长度为num的空列表
12         index,step=0,1
13         for x in s:
14             L[index] += x
15             if index == 0:
16                 step=1
17             elif index==numRows-1:
18                 step=-1
19             index += step
20         return ‘‘.join(L)
21 
22 s=‘PAYPALISHIRING‘
23 str=Solution.convert(1,s,3)
24 print(str)