String字符串形式的数学运算如何得到结果？
我用Scanner在控制台随意输入一段数学运算式，比如（3+5）*2
如何计算出结果？
有网友说：“用一个String类型的数组，将输入的字符串保存起来。然后对每一个char进行判断。如果要对所有的操作类型都判断，那么则涉及到对运算符号的配对以及优先级处理。如果只是题目中的一串，只需要用数组保存之后对每个字符进行判断即可。”
但是不知道怎么做，而且运算式是随机输入的，并不一定这么简单，可能有多个括号。
有哪位大神能帮忙解决这个问题，谢谢 java
[解决办法]
import java.util.Stack;
public class Evaluation {
public static void main(String[] args) {
// Check number of arguments passed
if (args.length != 1) {
System.out.println(
"Usage: java EvaluateExpression "expression"");
System.exit(1);
}
try {
System.out.println(evaluateExpression(args[0]) );
}catch (Exception ex) {
System.out.println("Wrong expression: " + args[0]);
}
}
/** Evaluate an expression */
public static int evaluateExpression(String expression) {
// Create operandStack to store operands
Stack<Integer> operandStack= new Stack<Integer>();
// Create operatorStack to store operators
Stack<Character> operatorStack= new Stack<Character>();
// Insert blanks around (, ), +, -, /, and *
expression = insertBlanks(expression);
// Extract operands and operators
String[] tokens = expression.split(" ");
// Phase 1: Scan tokens
for(String token : tokens){
if(token.length() == 0){
continue; // Back to the while loop to extract the next token
}else if(token.charAt(0) == '+'
[解决办法]
token.charAt(0) == '-'){
// Process all +, -, *, / in the top of the operator stack
while (!operatorStack.isEmpty() &&
(operatorStack.peek() == '+'
[解决办法]
operatorStack.peek() == '-'
[解决办法]
operatorStack.peek() == '*'
[解决办法]
operatorStack.peek() == '/')) {
processAnOperator(operandStack, operatorStack);
}
// Push the + or - operator into the operator stack
operatorStack.push(token.charAt(0));
}else if(token.trim().charAt(0) == '('){
operatorStack.push('('); //push '(' to stack
}else if(token.trim().charAt(0) == ')'){
while(operatorStack.peek() != '('){
processAnOperator(operandStack, operatorStack);
}
operatorStack.pop(); //Pop the '(' symbol from the stack
}else{ // An operand scanned
//push an operand to the stack
operandStack.push(new Integer(token));
}
}
// Phase 2: Process all the remaining operators in the stack
while(!operatorStack.isEmpty()){
processAnOperator(operandStack, operatorStack);
}
// Return the result
return operandStack.pop();
}
/** Process one operator: Take an operator from operatorStack and
apply it on the operands in the operandStack */
public static void processAnOperator(
Stack<Integer> operandStack, Stack<Character> operatorStack) {
char op = operatorStack.pop();
int op1 = operandStack.pop();
int op2 = operandStack.pop();
if (op == '+')
operandStack.push(op2 + op1);
else if (op == '-')
operandStack.push(op2 - op1);
else if (op == '*')
operandStack.push(op2 * op1);
else if (op == '/')
operandStack.push(op2 / op1);
}
public static String insertBlanks(String s) {
String result = "";
for (int i = 0; i < s.length(); i++) {
if (s.charAt(i) == '('
[解决办法]
s.charAt(i) == ')'
[解决办法]
s.charAt(i) == '+'
[解决办法]
s.charAt(i) == '-'
[解决办法]
s.charAt(i) == '*'
[解决办法]
s.charAt(i) == '/')
result += " " + s.charAt(i) + " ";
else{
result += s.charAt(i);
}
}
return result;
}
}
public static void main(String args[]) {
while (true) {
BufferedReader br = new BufferedReader(new InputStreamReader(
System.in));
System.out.print("请输入一个算式:");
try {
String str = br.readLine();
ScriptEngineManager mgr = new ScriptEngineManager();
ScriptEngine engine = mgr.getEngineByName("JavaScript");
Object o = engine.eval(str);
double d = Double.parseDouble(o.toString());
System.out.println(str + " = " + d);
} catch (Exception ex) {
ex.printStackTrace();
}
}
}
中缀表达式我没试过直接求值，但是应该也是可以实现的。
[解决办法]
直接用js的eval函数jdk 1.6后新增的调用脚本语言
import javax.script.Bindings;
import javax.script.ScriptEngine;
import javax.script.ScriptEngineManager;
import javax.script.ScriptException;
public class Test {
public static void main(String[] args) throws Exception {
ScriptEngineManager manager=new ScriptEngineManager();
ScriptEngine engine=manager.getEngineByExtension("js");
Bindings bindings=engine.createBindings();
String expression="(564-454)*10/2";
bindings.put("expression", expression);
Double value=(Double) engine.eval("eval(expression)", bindings);
System.out.println(value);
}
}