解析代表化学反应的字符串并验证反应是否可行 [英] Parsing a string that represents a chemical reaction and verify if the reaction is possible
问题描述
我必须编写一个程序,将用户的化学方程式作为输入,如12 CO2 + 6 H2O - > 2 C6H12O6 + 12 O2,并观察两个位点上的原子数量是否相同。有没有办法轻松计算和解析?
I have to write a program that takes a user's chemical equation as an input, like 12 CO2 + 6 H2O -> 2 C6H12O6 + 12 O2, and watch if the amount of Atoms is on both sites the same. Is there any way to calculate and parse this easily?
例如:
12 CO2 + 6 H2O - > 2 C6H12O6 + 12 O2
12 CO2 + 6 H2O -> 2 C6H12O6 + 12 O2
12 * 2 + 6 * 2 - > 2 * 6 + 2 * 12 + 2 * 6 + 12 * 2
12*2+6*2 -> 2*6+2*12+2*6+12*2
在这种情况下,应该输出false。
In this case there should be the Output "false".
这是我的代码,但它实际上只是尝试一些东西:
This is my code but it's actually is only to try out something:
public static void main(String[] args) {
Scanner s = new Scanner(System.in);
List<String> list = new ArrayList<String>();
String input = "";
while (!(input.equals("end"))) {
input = s.nextLine();
list.add(input);
}
list.remove(list.size() - 1);
for (int i = 0; i < list.size(); i++) {
int before = 0;
int after = 0;
String string = list.get(i);
string = besserUmwandeln(string);
System.out.println(string);
}
}
public static String besserUmwandeln(String string) {
string = string.replace("-", "");
string = string.trim().replaceAll(("\\s+"), " ");
string = string.replace(' ', '*');
StringBuilder builder = new StringBuilder(string);
System.out.println(string);
for (int k = 0; k < builder.length(); k++) {
if (Character.isUpperCase(builder.charAt(k))) {
builder.setCharAt(k, ':');
}
if (Character.isLowerCase(builder.charAt(k))) {
builder.setCharAt(k, '.');
}
if (Character.isDigit(builder.charAt(k))) {
} else {
}
}
for (int j = 0; j < builder.length(); j++) {
if (j < builder.length() && builder.charAt(j) == ':' && builder.charAt(j + 1) == '.') {
builder.deleteCharAt(j + 1);
}
}
for (int i = 0; i < builder.length(); i++) {
if (i < builder.length() - 1 && builder.charAt(i) == ':' && builder.charAt(i + 1) == ':') {
builder.deleteCharAt(i);
}
}
for (int i = 0; i < builder.length(); i++) {
if (i < builder.length() - 1 && builder.charAt(i) == '+' && builder.charAt(i + 1) == '*') {
builder.deleteCharAt(i + 1);
}
}
for (int i = 0; i < builder.length(); i++) {
if (i < builder.length() - 1 && builder.charAt(i) == '*' && builder.charAt(i + 1) == '+') {
builder.deleteCharAt(i);
}
}
for (int i = 0; i < builder.length(); i++) {
if (i < builder.length() - 1 && builder.charAt(i) == '*' && builder.charAt(i + 1) == '>') {
builder.deleteCharAt(i);
}
}
for (int i = 0; i < builder.length(); i++) {
if (i < builder.length() - 1 && builder.charAt(i) == '>' && builder.charAt(i + 1) == '*') {
builder.deleteCharAt(i + 1);
}
}
for (int i = 0; i < builder.length(); i++) {
if (i < builder.length() - 1 && builder.charAt(i) == '*' && builder.charAt(i + 1) == ':') {
builder.deleteCharAt(i + 1);
}
}
return builder.toString();
}
推荐答案
所以,每次我需要使用 Java
解析一些文本,我最终只使用 Regex
。所以我建议你也这样做。
So, every time I need to parse some text with Java
, I mostly end up just using Regex
. So I'd recommend you to also do so.
你可以在 regex101.com 。
并且还可以在 Java中轻松使用它
:
final inputText = ...
final Pattern pattern = Patern.compile("Some regex code");
final Matcher matcher = pattern.matcher(input);
if (matcher.find()) {
System.out.println(matcher.group(0));
}
内部正则表达式
你可以使用(
和)
定义捕获组,然后通过 matcher.group获取结果( int)
。
Inside Regex
you can define capturing groups with (
and )
and then grab the results by matcher.group(int)
.
例如,您可以先使用(。*) - >分隔等式。 (。*)
。
然后使用 find
循环左右组:(\d +)(\ w +)(?:\ + | - | $)
。
Then loop the left and right group using find
with: (\d+) (\w+)(?: \+| -|$)
.
之后,您可以使用 group(1)
获取金额,使用 group(2)
获取元素。
After that you can use group(1)
for the amount and group(2)
for the element.
如果需要,还可以使用(\w)(\\\?)来迭代第二组(元素)以获得精确的元素分布
。然后第一组是元素,例如对于文本 CO2
它产生两次点击,第一次点击具有 group(1) - > C
而没有第二组。第二个点击 group(1) - > O
和 group(2) - > 2
。
And if needed also iterate the second group (the element) for the exact element distribution using (\w)(\d?)
. Then the first group is the element, for example for the text CO2
it yields two hits, the first hit has group(1) -> C
and no second group. The second hit has group(1) -> O
and group(2) -> 2
.
在此测试您的正则表达式: regex101#Q6KMJo
Test your regex here: regex101#Q6KMJo
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