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Difficulty: Medium
You are given a string s consisting of lowercase English letters and the special characters: *, #, and %.
Build a new string result by processing s according to the following rules from left to right:
result.* removes the last character from result, if it exists.# duplicates the current result and appends it to itself.% reverses the current result.Return the final string result after processing all characters in s.
s = "a#b%*"
"ba"
| i | s[i] | Operation | Current Result |
|---|---|---|---|
| 0 | 'a' | Append 'a' | "a" |
| 1 | '#' | Duplicate result | "aa" |
| 2 | 'b' | Append 'b' | "aab" |
| 3 | '%' | Reverse result | "baa" |
| 4 | '*' | Remove the last character | "ba" |
Thus, the final result is:
"ba"
s = "z*#"
""
| i | s[i] | Operation | Current Result |
|---|---|---|---|
| 0 | 'z' | Append 'z' | "z" |
| 1 | '*' | Remove the last character | "" |
| 2 | '#' | Duplicate the string | "" |
Thus, the final result is:
""
1 <= s.length <= 20
s consists of only lowercase English letters and special characters *, #, and %.
import java.util.*;
public class Main {
public static String processStr(String s) {
StringBuilder result = new StringBuilder();
for (char ch : s.toCharArray()) {
if (ch >= 'a' && ch <= 'z') {
result.append(ch);
}
else if (ch == '*') {
if (result.length() > 0) {
result.deleteCharAt(result.length() - 1);
}
}
else if (ch == '#') {
result.append(result.toString());
}
else if (ch == '%') {
result.reverse();
}
}
return result.toString();
}
public static void main(String[] args) {
String s = "a#b%*";
System.out.println(processStr(s));
}
}
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Drag and arrange the algorithm steps in the correct execution order instead of spending time typing code letter by letter.
The algorithm is divided into three logical parts. Carefully rearrange each section in the correct order to form a complete and valid solution.
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