Java > Recursion-1 > parenBit
prev  |  next  |  chance

Given a string that contains a single pair of parenthesis, compute recursively a new string made of only of the parenthesis and their contents, so "xyz(abc)123" yields "(abc)". parenBit("xyz(abc)123") → "(abc)" parenBit("x(hello)") → "(hello)" parenBit("(xy)1") → "(xy)" ...Save, Compile, Run public String parenBit(String str) { } See also Java Example Code. Java help docs: If Boolean Logic | Strings | While and For Loops | Arrays and Loops prev  |  next  |  chance   |  CodingBat  >  Recursion-1

Forget It! -- delete my code for this problem 378.0

Copyright Nick Parlante 2006-10 - privacy