标签:arm pre function stack throws illegal ase pen static
package DataStructures.Stacks; import java.util.Stack; public class DecimalToAnyUsingStack { public static void main(String[] args) { assert convert(0, 2).equals("0"); assert convert(30, 2).equals("11110"); assert convert(30, 8).equals("36"); assert convert(30, 10).equals("30"); assert convert(30, 16).equals("1E"); } /** * Convert decimal number to another radix * * @param number the number to be converted * @param radix the radix * @return another radix * @throws ArithmeticException if <tt>number</tt> or <tt>radius</tt> is invalid */ private static String convert(int number, int radix) { if (radix < 2 || radix > 16) { throw new ArithmeticException( String.format("Invalid input -> number:%d,radius:%d", number, radix)); } char[] tables = { ‘0‘, ‘1‘, ‘2‘, ‘3‘, ‘4‘, ‘5‘, ‘6‘, ‘7‘, ‘8‘, ‘9‘, ‘A‘, ‘B‘, ‘C‘, ‘D‘, ‘E‘, ‘F‘ }; Stack<Character> bits = new Stack<>(); do { bits.push(tables[number % radix]); number = number / radix; } while (number != 0); StringBuilder result = new StringBuilder(); while (!bits.isEmpty()) { result.append(bits.pop()); } return result.toString(); } }
这个就是利用栈来进行进制转换
package DataStructures.Stacks; import java.util.Stack; /** * The nested brackets problem is a problem that determines if a sequence of * brackets are properly nested. A sequence of brackets s is considered properly * nested if any of the following conditions are true: - s is empty - s has the * form (U) or [U] or {U} where U is a properly nested string - s has the form * VW where V and W are properly nested strings For example, the string * "()()[()]" is properly nested but "[(()]" is not. The function called * is_balanced takes as input a string S which is a sequence of brackets and * returns true if S is nested and false otherwise. * * @author akshay sharma * @author <a href="https://github.com/khalil2535">khalil2535<a> * @author shellhub */ class BalancedBrackets { /** * Check if {@code leftBracket} and {@code rightBracket} is paired or not * * @param leftBracket left bracket * @param rightBracket right bracket * @return {@code true} if {@code leftBracket} and {@code rightBracket} is paired, * otherwise {@code false} */ public static boolean isPaired(char leftBracket, char rightBracket) { char[][] pairedBrackets = { {‘(‘, ‘)‘}, {‘[‘, ‘]‘}, {‘{‘, ‘}‘}, {‘<‘, ‘>‘} }; for (char[] pairedBracket : pairedBrackets) { if (pairedBracket[0] == leftBracket && pairedBracket[1] == rightBracket) { return true; } } return false; } /** * Check if {@code brackets} is balanced * * @param brackets the brackets * @return {@code true} if {@code brackets} is balanced, otherwise {@code false} */ public static boolean isBalanced(String brackets) { if (brackets == null) { throw new IllegalArgumentException("brackets is null"); } Stack<Character> bracketsStack = new Stack<>(); for (char bracket : brackets.toCharArray()) { switch (bracket) { case ‘(‘: case ‘[‘: case ‘{‘: bracketsStack.push(bracket); break; case ‘)‘: case ‘]‘: case ‘}‘: if (bracketsStack.isEmpty() || !isPaired(bracketsStack.pop(), bracket)) { return false; } break; default: /* other character is invalid */ return false; } } return bracketsStack.isEmpty(); } public static void main(String[] args) { assert isBalanced("[()]{}{[()()]()}"); assert !isBalanced("[(])"); } }
匹配括号
比较简单,今天主要在学习函数式编程。
标签:arm pre function stack throws illegal ase pen static
原文地址:https://www.cnblogs.com/CherryTab/p/11964307.html