替代Java中嵌套While循环的方法

huangapple
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英文:

Alternative way to replace Nested While Loop in Java

问题

我尝试在Java中使用以下代码计算数学表达式：

  1. public double applyOp(char op, double b, double a) {
  2.     switch (op) {
  3.         case '+':
  4.             return a + b;
  5.         case '-':
  6.             return a - b;
  7.         case '*':
  8.             return a * b;
  9.         case '/':
  10.             return a / b;
  11.     }
  12.     return 0;
  13. }
  15. public boolean hasPrecedence(char op1, char op2) {
  16.     return (op1 != '*' && op1 != '/') || (op2 != '+' && op2 != '-');
  17. }
  19. public double evaluate(String input) {
  20.     Stack<Double> values = new Stack<>();
  21.     Stack<Character> ops = new Stack<>();
  22.     int stringIndex = 0;
  24.     while (stringIndex < input.length()) {
  25.         StringBuilder multiDigitsNumber = new StringBuilder();
  27.         // If the input is a number, put it onto the values stack
  28.         if (input.charAt(stringIndex) >= '0' && input.charAt(stringIndex) <= '9') {
  29.             while (stringIndex < input.length() && input.charAt(stringIndex) >= '0' && input.charAt(stringIndex) <= '9') {
  30.                 multiDigitsNumber.append(input.charAt(stringIndex++));
  31.             }
  32.             values.push(Double.parseDouble(multiDigitsNumber.toString()));
  33.         }
  35.         // If the input is an operator, put it onto the ops stack
  36.         else {
  37.             while (!ops.empty() && hasPrecedence(input.charAt(stringIndex), ops.peek())) {
  38.                 values.push(applyOp(ops.pop(), values.pop(), values.pop()));
  39.             }
  40.             ops.push(input.charAt(stringIndex++));
  41.         }
  42.     }
  44.     // Execute remaining operators in the values stack
  45.     while (!ops.empty()) {
  46.         values.push(applyOp(ops.pop(), values.pop(), values.pop()));
  47.     }
  49.     // The final number in the values stack is the result
  50.     return values.pop();
  51. }
  53. Input example:
  55. > 12+24*2-30/5.....

上述代码正常工作，但我想知道是否有方法可以替代下面这部分：

  1. while (stringIndex < input.length() && input.charAt(stringIndex) >= '0' && input.charAt(stringIndex) <= '9') {
  2.     multiDigitsNumber.append(input.charAt(stringIndex++));
  3. }

以避免在这种情况下使用嵌套的 while 循环。目标是在字符串中捕获数字，直到遇到运算符。提前感谢您的帮助。

英文:

I tried to Evaluate Mathematical Expressions in Java with the following code:

  1. public double applyOp(char op,double b,double a)
  2. {
  3. switch (op)
  4. {
  5. case &#39;+&#39;:
  6. return a + b;
  7. case &#39;-&#39;:
  8. return a - b;
  9. case &#39;*&#39;:
  10. return a * b;
  11. case &#39;/&#39;:
  12. return a / b;
  13. }
  14. return 0;
  15. }
  16. public boolean hasPrecedence(char op1,char op2)
  17. {
  18. return (op1 != &#39;*&#39; &amp;&amp; op1 != &#39;/&#39;) || (op2 != &#39;+&#39; &amp;&amp; op2 != &#39;-&#39;);
  19. }
  20. public double evaluate(String input) {
  21. Stack&lt;Double&gt; values = new Stack&lt;&gt;();
  22. Stack&lt;Character&gt; ops = new Stack&lt;&gt;();
  23. int stringIndex = 0;
  24. while (stringIndex &lt; input.length())
  25. {
  26. StringBuilder multiDigitsNumber = new StringBuilder();
  27. // If the input is number put to stack values
  28. if (input.charAt(stringIndex) &gt;= &#39;0&#39; &amp;&amp; input.charAt(stringIndex) &lt;= &#39;9&#39;)
  29. {
  30. while (stringIndex &lt; input.length() &amp;&amp; input.charAt(stringIndex) &gt;= &#39;0&#39; &amp;&amp; input.charAt(stringIndex) &lt;= &#39;9&#39;)
  31. {
  32. multiDigitsNumber.append(input.charAt(stringIndex++));
  33. }
  34. values.push(Double.parseDouble(multiDigitsNumber.toString()));
  35. }
  36. // If the input is operator put to stack ops
  37. else
  38. {
  39. while (!ops.empty() &amp;&amp; hasPrecedence(input.charAt(stringIndex),ops.peek()))
  40. {
  41. values.push(applyOp(ops.pop(),values.pop(),values.pop()));
  42. }
  43. ops.push(input.charAt(stringIndex++));
  44. }
  45. }
  46. // Execute remain operator in stack values
  47. while (!ops.empty()) {
  48. values.push(applyOp(ops.pop(), values.pop(), values.pop()));
  49. }
  50. // The final number in stack value is result
  51. return values.pop();
  52. }

Input example:

> 12+24*2-30/5.....

The code above works fine but I wonder are there any way to replace

  1. while (stringIndex &lt; input.length() &amp;&amp; input.charAt(stringIndex) &gt;= &#39;0&#39; &amp;&amp; input.charAt(stringIndex) &lt;= &#39;9&#39;)
  2. {
  3. multiDigitsNumber.append(input.charAt(stringIndex++));
  4. }

with something else so I don't have to use nested while loop in this situation. The goal is to catch number in string until it reach an operator
Thanks in advance

答案1

得分: 1

可以像这样使用正则表达式（Regex）。

  1. if (input.charAt(stringIndex) >= '0' && input.charAt(stringIndex) <= '9')
  2. {
  3.     String number = input.substring(stringIndex).replaceAll("^(\\d+).*", "$1");
  4.     values.push(Double.parseDouble(number));
  5.     stringIndex += number.length();
  6. }
英文:

You can use Regex like this.

  1. if (input.charAt(stringIndex) &gt;= &#39;0&#39; &amp;&amp; input.charAt(stringIndex) &lt;= &#39;9&#39;)
  2. {
  3. String number = input.substring(stringIndex).replaceAll(&quot;^(\\d+).*&quot;, &quot;$1&quot;);
  4. values.push(Double.parseDouble(number));
  5. stringIndex += number.length();
  6. }

huangapple
  • 本文由 发表于 2020年8月27日 11:55:28
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  • java
  • nested-loops
  • while-loop
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