java解压缩包含更多嵌套字符串的字符串
赋值包括对字符串进行解压缩。特别是,代码必须适用于3个示例,如图所示
我这里的代码在前两个示例中有效。然而,我不能拿出第三个样本。可能我不理解递归的概念。你能帮我吗
import java.util.Scanner;
public class Compression4 {
public static void main(String[] args)
{
Scanner in = new Scanner(System.in);
String input=in.next();
System.out.println(uncompress(input));
}
public static boolean flag = true;
public static String uncompress(String compressedText)
{
return uncompress(compressedText, "", "");
}
public static String getMultiple(String x, int N) {
if (N == 0) return "";
return ""+x+getMultiple(x,N-1);
}
public static String uncompress(String text, String count, String output)
{
if (text.equals(""))
{
return output;
}
if(text.charAt(0) == '(')
{
int FirstIndex = text.indexOf("(")+1;
String inner = text.substring(FirstIndex, text.lastIndexOf(")"));
//System.out.println(inner);
flag = false;
return uncompress (inner, count, output);
}
else if (Character.isLetter(text.charAt(0)))
{
//letter case - need to take the count we have accrued, parse it into an integer and add to output
if (flag==true)
{
//System.out.println(count);// * text.charAt(0);
String s = String.valueOf(text.charAt(0));
output += getMultiple(s,Integer.parseInt(count));
count ="1";
}
else
{
//System.out.println(count);// * text.charAt(0);
output += getMultiple(text,Integer.parseInt(count));
//System.out.println("output: "+output);
count="0";
}
}
else if(Character.isDigit(text.charAt(0)))
{
//digit case - need to add to the count but keep as a string because must be parsed later
if(flag)
count += (""+text.charAt(0));
else
{
count = "0";
count += (""+text.charAt(0));
}
}
//parse the *remainder* of the string, one character at a time, so pass in the substring(1)
return uncompress(text.substring(1), count, output);
}
}
# 1 楼答案
抱歉,代码太长了,但用代码解释比用文字解释更容易
前提:
算法阶段:
递归:逻辑基于语言的语法。递归的关键概念:
我已经做了很多评论来解释这个算法在做什么。如果不清楚,我可以更好地解释