Java 字符串计算器
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Java String Calculator
提问by tmax
I'm trying to make a simple calculator in Java which takes input in the form of a string and does a simple '+' and '-' operation.
我正在尝试用 Java 制作一个简单的计算器，它以字符串的形式输入并执行简单的“+”和“-”操作。
Single digit inputs work but my problem is when i try to implement this for double digit
一位数输入有效，但我的问题是当我尝试为两位数实现这个时
input string is: 5+20+5+11
list 1 = [5, 20, 2, 0, 5, 11, 1]
list 2 = [+, +, +]
Answer:27
输入字符串为：5+20+5+11
list 1 = [5, 20, 2, 0, 5, 11, 1]
list 2 = [+, +, +]
答案：27
I need to find a way where after storing [5] in list1 how i can add [5,20] instead of [5,20,2,0] which the current code is doing.
我需要找到一种方法，在将 [5] 存储在 list1 中之后，我如何添加 [5,20] 而不是当前代码正在执行的 [5,20,2,0] 。
public int calC(String input) {
int len = input.length();
ArrayList list1 = new ArrayList();
ArrayList list2 = new ArrayList();
for (int i = 0; i < len; i++) {
if ((input.charAt(i) != '+') && (input.charAt(i) != '-')) {
// check if the number is double-digit
if ((i + 1 <= len - 1)) {
if ((input.charAt(i + 1) != '+')&& (input.charAt(i + 1) != '-')) {
String temp = "";
temp = temp + input.charAt(i) + input.charAt(i + 1);
int tempToInt = Integer.parseInt(temp);
// adding the double digit number
list1.add(tempToInt);
}
// add single digit number
list1.add(input.charAt(i) - '0');
}
} else {
// adding the symbols
list2.add(input.charAt(i));
}
}
int result = 0;
result = result + (int) list1.get(0);
for (int t = 0; t < list2.size(); t++) {
char oper = (char) list2.get(t);
if (oper == '+') {
result = result + (int) list1.get(t + 1);
} else if (oper == '-') {
result = result - (int) list1.get(t + 1);
}
}
return result;
}
Edit: working version
@Ker p pag thanks for the updated methods
input string is: 5+20+5+11
[5, 20, 5, 11]
[+, +, +]
Answer:41
I'll need to try to implement this with stack as suggested but the current version works
编辑：工作版本
@Ker p pag 感谢更新方法
输入字符串是：5+20+5+11
[5, 20, 5, 11]
[+, +, +]
答案：41
我需要尝试按照建议使用堆栈实现此功能，但当前版本有效
static boolean isDigit(char check) {
if (Character.isDigit(check)) {
return true;
}
return false;
}
public static int calC(String input) {
int len = input.length();
ArrayList list1 = new ArrayList();
ArrayList list2 = new ArrayList();
for (int i = 0; i < len; i++) {
if ((i + 1 <= len - 1)) {
if (isDigit(input.charAt(i)) && isDigit(input.charAt(i + 1))) {
String temp = input.charAt(i) + "" + input.charAt(i + 1);
int toInt = Integer.parseInt(temp);
list1.add(toInt);
i = i+1;
} else if (isDigit(input.charAt(i))) {
list1.add(input.charAt(i)- '0');
} else {
list2.add(input.charAt(i));
}
}
}
int result = 0;
result = result + (int) list1.get(0);
for (int t = 0; t < list2.size(); t++) {
char oper = (char) list2.get(t);
if (oper == '+') {
result = result + (int) list1.get(t + 1);
} else if (oper == '-') {
result = result - (int) list1.get(t + 1);
}
}
return result;
}
回答by Ker p pag
for storing the input to the list you could try this snippet
要将输入存储到列表中，您可以尝试此代码段
for (int i = 0; i < input.length() - 1; i++) {
// make a method
// check if current character is number || check if current
// character is number and the next character
if (isDigit(input.charAt(i)) && isDigit(input.charAt(i + 1))) {
list.add(input.charAt(i) +""+ input.charAt(i + 1));
} else if (isDigit(input.charAt(i))) {
list.add(input.charAt(i));
}else{
operator.add(input.charAt(i));
}
}
//check if it is a number
public boolean isDigit(char input){
if(input == '1' ||
input == '2' ||
input == '3' ||
input == '4' ||
input == '5' ||
input == '6' ||
input == '7' ||
input == '8' ||
input == '9' ||
input == '0')
return true;
return false;
}
回答by Wundwin Born
If you want the result 41for input string "5+20+5+11",
why not use ScriptEngineManagerwith JavaScriptengine,
如果您想要41输入 string的结果"5+20+5+11"，为什么不ScriptEngineManager与JavaScript引擎一起使用，
public double calC(String input) {
int result = 0;
ScriptEngineManager mgr = new ScriptEngineManager();
ScriptEngine engine = mgr.getEngineByName("JavaScript");
return (Double)engine.eval(input);
}
But note that the return type is doublehere.
但请注意，返回类型在double这里。
If you want only intas return type in this case, try with this
如果你只想要int在这种情况下作为返回类型，试试这个
return new BigDecimal(engine.eval(input).toString()).intValue();
回答by Puneet_Techie
I agree that stack is the best solution,but still giving an alternative way of doing this.
我同意堆栈是最好的解决方案，但仍然提供了一种替代方法。
String input = "5+20+11+1";
StringBuilder sb = new StringBuilder();
List<Integer> list1 = new ArrayList<Integer>();
List<Character> list2 = new ArrayList<Character>();
char[] ch = input.toCharArray();
for(int i=0;i<ch.length;i++)
{
if(ch[i]!='+')
{
sb.append(ch[i]);
}else
{
list2.add(ch[i]);
list1.add(Integer.valueOf(sb.toString()));
sb.setLength(0);
}
}
if(sb.length()!=0)
list1.add(Integer.valueOf(sb.toString()));
System.out.println(list1.size());
for(Integer i:list1)
{
System.out.println("values"+i);
}
回答by AlexanderBrevig
Another way to think about this:
另一种思考方式：
public class InlineParsing {
public static void main(String []args){
String input = "5-2+20+5+11-10";
input = input.replace(" ","");
String parsedInteger = "";
String operator = "";
int aggregate = 0;
for (int i = 0; i < input.length(); i++){
char c = input.charAt(i);
if (Character.isDigit(c)) {
parsedInteger += c;
}
if (!Character.isDigit(c) || i == input.length()-1){
int parsed = Integer.parseInt(parsedInteger);
if (operator == "") {
aggregate = parsed;
}
else {
if (operator.equals("+")) {
aggregate += parsed;
}else if (operator.equals("-")){
aggregate -= parsed;
}
}
parsedInteger ="";
operator = ""+c;
}
}
System.out.println("Sum of " + input+":\r\n" + aggregate);
}
}
It's basically a state machine that traverses over each char.
它基本上是一个遍历每个字符的状态机。
Iterate over each char:
迭代每个字符：
- if current char is a digit, add to current number buffer
- if current char is not a digit or we're parsing the last digit
- if an operator has been parsed use that to add the newly parsed number to the sum if no operator has been parsed, set sum to current parsed number
- clear current number buffer
- store current char as operator
- 如果当前字符是数字，则添加到当前数字缓冲区
- 如果当前字符不是数字或者我们正在解析最后一位数字
- 如果已经解析了运算符，则使用该运算符将新解析的数字添加到总和中，如果未解析任何运算符，则将总和设置为当前已解析的数字
- 清除当前号码缓冲区
- 将当前字符存储为运算符
回答by HJK
Here is the code:
这是代码：
String a = "5+20-15+8";
System.out.println(a);
String operators[]=a.split("[0-9]+");
String operands[]=a.split("[+-]");
int agregate = Integer.parseInt(operands[0]);
for(int i=1;i<operands.length;i++){
if(operators[i].equals("+"))
agregate += Integer.parseInt(operands[i]);
else
agregate -= Integer.parseInt(operands[i]);
}
System.out.println(agregate);
回答by chAlexey
I could advise you to use Exp4j. It is easy to understand as you can see from the following example code:
我可以建议您使用Exp4j。从以下示例代码中可以看出，它很容易理解：
Expression e = new ExpressionBuilder("3 * sin(y) - 2 / (x - 2)")
.variables("x", "y")
.build()
.setVariable("x", 2.3)
.setVariable("y", 3.14);
double result = e.evaluate();
Especially for the case of using more complex expression this could be a better choice.
特别是对于使用更复杂表达式的情况，这可能是更好的选择。
回答by Abhishek Singh
private static int myCal() {
String[] digits = {
"1",
"2",
"3",
"4",
"5"
};
String[] ops = {
"+",
"+",
"+",
"-"
};
int temp = 0;
int res = 0;
int count = ops.length;
for (int i = 0; i < digits.length; i++) {
res = Integer.parseInt(digits[i]);
if (i != 0 && count != 0) {
count--;
switch (ops[i - 1]) {
case "+":
temp = Math.addExact(temp, res);
break;
case "-":
temp = Math.subtractExact(temp, res);
break;
case "*":
temp = Math.multiplyExact(temp, res);
break;
case "/":
temp = Math.floorDiv(temp, res);
break;
}
}
}
return temp;
}
