parsing - Match brackets the kotlin way -


i'm giving kotlin go; coding contently, have arraylist of chars want classify depending on how brackets matched:

(abcde)  // ok characters other brackets can go anywhere )abcde(  // ok matching brackets 'invertedly' ok (({()})) // ok  )()()([] // ok  ([)]     // bad can't have different overlapping bracket pairs (((((    // bad brackets need have match

my solution comes out(recursive):

//charlist property //recursion starter'upper private fun classifylistofcharacters() : boolean{     var j = 0     while (j < charlist.size ) {         if (charlist[j].isbracket()){             j = checkmatchingbrackets(j+1, charlist[j])         }         j++     }     return j == commandlist.size }  private fun checkmatchingbrackets(i: int, firstbracket :char) : int{     var j =     while (j < charlist.size ) {         if (charlist[j].isbracket()){             if (charlist[j].matchesbracket(firstbracket)){                  return j //matched bracket normal/inverted             }             j = checkmatchingbrackets(j+1, charlist[j])         }         j++     }     return j }

this works, how in kotlin? feels i've coded java in kotlin syntax

found functional languages better @ recursion, i've tried thinking in terms of manipulating functions , sending them down recursion no avail. i'd glad pointed in right direction, code, or pseudo-code of possible refactoring.

(omitted extension methods regarding brackets, think it's clear do)

another, possibly simpler approach problem maintaining stack of brackets while iterate on characters.

when encounter bracket:

  • if matches top of stack, pop top of stack;

  • if not match top of stack (or stack empty), push onto stack.

if brackets remain on stack @ end, means unmatched, , answer false. if stack ends empty, answer true.

this correct, because bracket @ position i in sequence can match 1 @ position j, if there's no unmatched bracket of different kind between them (at position k, i < k < j). stack algorithm simulates logic of matching.

basically, algorithm implemented in single for-loop:

val stack = stack<char>()  (c in charlist) {     if (!c.isbracket())         continue      if (stack.isnotempty() && c.matchesbracket(stack.peek())) {         stack.pop()     } else {         stack.push(c)     } }  return stack.isempty()

i've reused extensions c.isbracket(...) , c.matchesbracket(...). stack<t> jdk class.

this algorithm hides recursion , brackets nesting inside abstraction of brackets stack. compare: current approach implicitly uses function call stack instead of brackets stack, purpose same: either finds match top character or makes deeper recursive call character on top.


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