leetcode刷题笔记六 Z字型转换 Scala版本
源地址:(leetcode刷题笔记六 Z字型转换 Scala版本)[https://leetcode.com/problems/zigzag-conversion/submissions/]
问题描述:
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);
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
简要思路分析:
设置层数指示器与方向参数,通过判断执行是否达到顶部或者底部,进行方向变更,将对应层的字符放入对应的字符串数组中,最后将字符串数组进行汇总。
代码补充:
object Solution {
def convert(s: String, numRows: Int): String = {
if (numRows == 1 || s.length <= numRows) return s
var upDown = true
var counter = 0
var arr:Array[String] = new Array[String](numRows)
var answer = ""
for (i <- 0 until s.length){
//println("--------------------------")
//println("i: " + i.toString)
//println("counterPre: "+counter.toString)
//println("upDownPre: "+upDown.toString)
arr(counter) += s.charAt(i)
/* 用match不可以。。 神秘
(counter) match{
case(0) => upDown = true
case(maxCounter) => upDown = false
}
*/
if (counter == 0) upDown = true
if (counter == numRows-1) upDown = false
(upDown) match{
case(true) => counter+=1
case(false) => counter-=1
}
//println("counterAfter: "+counter.toString)
//println("upDownAfter: "+upDown.toString)
//println("s(i): " + s.charAt(i))
}
for (i <- 0 until numRows){
answer += arr(i).drop(4).toString
//println(arr(i))
}
return answer
}
}