後綴表達式輸出中缺少空格 - Java

我的代碼成功地將中綴表達式轉換爲後綴表達式。但是，當我輸入的數字超過1位數時（例如546），我在這個數字和右邊的操作數之間沒有空格。後綴表達式輸出中缺少空格 - Java

我的代碼的測試運行：輸入：輸入表達式：（24/4）/（15/3）* 10 - 4 + 2 輸出：後綴表達式爲：244/153// 10 * 4 - 2+

我想它是後綴表達式是：24 4/15 3// 10 * 4 - 2+

這是我的代碼：請建議，讓我的任何變化在輸出中插入空格。

import java.util.*; 

public class PostfixConversion { 

    public static void main(String args[]) { 

     System.out.print("Enter an expression: "); 
     String infix = new Scanner(System.in).nextLine(); 

     System.out.println(convertToPostfix(infix)); 

    } 


    public static boolean precedence(char first, char second) 
    { 
     int v1 = 0, v2 = 0; 
     //find value for first 
     if(first == '-' || first == '+'){ 
     v1 = 1; 
     }else if(first == '*' || first == '/'){ 
     v1 = 2;  
     }//end if 

     //find value for second 
     if(second == '-' || second == '+'){ 
     v2 = 1; 
     }else if(second == '*' || second == '/'){ 
     v2 = 2;  
     }//end if 

    if(v1 < v2){ 
     return false; 
    }//end if 

    return true; 
    }//end precedence method 

//converts infix expression into postfix expression 
public static String convertToPostfix(String infixExp) 
    { 
    String postFix = "The Postfix Expression is: "; 
    Stack<Character> stack = new Stack<Character>(); 
    char character = ' '; 

    for(int i = 0; i < infixExp.length(); i++) 
    { 
     character = infixExp.charAt(i); 

     //determine if character is an operator 
     if(character == '*' || character == '-' || character == '/' || character == '+') 
     { 
      while(!stack.empty() && precedence(stack.peek(), character)){ 
       postFix += stack.pop(); 
      }//end while 
      stack.push(character); 
     } 
     else if(character == '(') //check for left parenthesis 
     { 
      stack.push(character); 
     } 
     else if (character == ')') 
     { 
      while(!stack.peek().equals('(') && !stack.isEmpty()){ //add characters until left parenthesis 
       postFix += stack.pop(); 
      }//end while 

      if(!stack.isEmpty() && stack.peek().equals('(')){ 
       stack.pop(); // pop/remove left parenthesis 
      } 
     } 
     else 
     { 
      postFix += character; 
     }//end if 
    }//end for 
    while(!stack.empty()) //add the remaining elements of stack to postfix expression 
    { 
     if(stack.peek().equals('(')) 
     { 
      postFix = "There is no matching right parenthesis."; 
      return postFix; 
     } 
     postFix += stack.pop(); 
    } 
     return postFix; 
}//end convertToPo 
}

來源

2017-10-21 Bharat Edupghanti

[什麼是調試器，它如何幫助我診斷問題？]（https://stackoverflow.com/q/25385173/5221149） – Andreas

查找Djkstra分流碼算法的一些現有實現，以查看'precedence（）'方法應該是真的實現。它並沒有你所做的那麼複雜。 – EJP

的解決辦法是在這裏添加一行（見代碼註釋）：

if(character == '*' || character == '-' || character == '/' || character == '+') 
{ 
    postFix += ' '; // <-- add space here 
    while(!stack.empty() && precedence(stack.peek(), character)){ 
     postFix += stack.pop(); 
    }//end while 
    stack.push(character); 
}

應用在我的機器看起來像固定輸出後：

The Postfix Expression is: 24 4/ 15 3//10 * 4 - 2+

來源

2017-10-21 23:38:48 alfasin

後綴表達式輸出中缺少空格 - Java

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