嘿,我目前正在用Java编写一个表达式计算器。我现在已经让它工作了,但无法弄清楚如何在终端中间隔写我想要的是两个相邻的列表。我最初会使用选项卡,但由于数组可以增长,它会移动另一个选项卡。这是我现在得到的:
OUTPUT BUFFER STACK
[] []
[(] []
[(] [4]
[(, +] [4]
[(, +, (] [4]
[(, +, (] [4, 2]
[(, +, (, *] [4, 2]
[(, +, (, *, (] [4, 2]
[(, +, (, *, (] [4, 2, -2]
[(, +, (, *, (, -] [4, 2, -2]
[(, +, (, *, (, -] [4, 2, -2, 1]
[(, +, (, *] [4, 2, -2, 1, -]
[(, +] [4, 2, -2, 1, -, *]
[] [4, 2, -2, 1, -, *, +]
[*] [4, 2, -2, 1, -, *, +]
[*] [4, 2, -2, 1, -, *, +, 1.5]
[%] [4, 2, -2, 1, -, *, +, 1.5, *]
FINAL OUTPUT: [4, 2, -2, 1, -, *, +, 1.5, *, 2, %]
INFIX: ( 4 + ( 2 * ( -2 - 1 ) ) ) * 1.5 % 2
POSTFIX: 4 2 -2 1 - * + 1.5 * 2 %
TOKEN ACTION STACK
4 Add [4.0]
2 Add [4.0, 2.0]
-2 Add [4.0, 2.0, -2.0]
1 Add [4.0, 2.0, -2.0, 1.0]
- Operate [4.0, 2.0, -3.0]
* Operate [4.0, -6.0]
+ Operate [-2.0]
1.5 Add [-2.0, 1.5]
* Operate [-3.0]
2 Add [-3.0, 2.0]
% Operate [-1.0]
ANSWER: -1.0
以下是我想要实现的目标:
OUTPUT BUFFER STACK
[] []
[(] []
[(] [4]
[(, +] [4]
[(, +, (] [4]
[(, +, (] [4, 2]
[(, +, (, *] [4, 2]
[(, +, (, *, (] [4, 2]
[(, +, (, *, (] [4, 2, -2]
[(, +, (, *, (, -] [4, 2, -2]
[(, +, (, *, (, -] [4, 2, -2, 1]
[(, +, (, *] [4, 2, -2, 1, -]
[(, +] [4, 2, -2, 1, -, *]
[] [4, 2, -2, 1, -, *, +]
[*] [4, 2, -2, 1, -, *, +]
[*] [4, 2, -2, 1, -, *, +, 1.5]
[%] [4, 2, -2, 1, -, *, +, 1.5, *]
FINAL OUTPUT: [4, 2, -2, 1, -, *, +, 1.5, *, 2, %]
INFIX: ( 4 + ( 2 * ( -2 - 1 ) ) ) * 1.5 % 2
POSTFIX: 4 2 -2 1 - * + 1.5 * 2 %
TOKEN ACTION STACK
4 Add [4.0]
2 Add [4.0, 2.0]
-2 Add [4.0, 2.0, -2.0]
1 Add [4.0, 2.0, -2.0, 1.0]
- Operate [4.0, 2.0, -3.0]
* Operate [4.0, -6.0]
+ Operate [-2.0]
1.5 Add [-2.0, 1.5]
* Operate [-3.0]
2 Add [-3.0, 2.0]
% Operate [-1.0]
ANSWER: -1.0
这是我当前用于打印堆栈和输出缓冲区的代码:
System.out.println("OUTPUT BUFFERttttSTACK");
ArrayList<String> out = new ArrayList<String>();
Stack<String> stack = new Stack<String>();
for(String token : inputTokens){
System.out.print(stack+ "ttttt");
System.out.println(out);
if(isOperator(token)){
while(!stack.empty() && isOperator(stack.peek())){
if (checkPri(token, stack.peek())<=0 || checkPri(token, stack.peek()) < 0 ){
out.add(stack.pop());
//System.out.println("OUTPUT: " + out);
continue;
}
break;
}
stack.push(token);
}else if(token.equals("(")){
stack.push(token);
}else if(token.equals(")")){
while(!stack.empty() && !stack.peek().equals("(")){
out.add(stack.pop());
}
stack.pop();
} else {
out.add(token);
//System.out.println("OUTPUT BUFFER: " + out);
}
}
您可以使用String.format(String,args...
)或System.out.printf(String,args...);
由于我的英语不好,我不知道如何解释,但这里有一个例子:
int i=10000,j=200,k=5000000;
System.out.printf("%10d%10d%10d",i,j,k);
将获得以下输出:
10000 200 5000000
^ ^ ^each has ten spaces between them because of %10d