我无法正确表达我的问题,所以请耐心等待。举个例子可能会更好。
这是我参加 P&Q&R 时得到的结果
PQR:
TTT = T
FFF = F
TTF = F
FFT = T
TFF = F
FTT = T
TFT = T
FTF = F
它只是查看 R 并从那里获取值,P&Q 似乎被完全忽略了。如果我为 P&Q&~R 运行,值将被反转,但如果我运行例如 ~P&~Q&R,值将保持不变。
目前我的代码太长了,但我正在努力缩短它以避免重复代码。不过,目前我只想解决这个问题。
这是应该评估 FTF 的代码片段:
//Evaluate PQR = false, true, false, ftf
for(int i=0;i<x.length();i++)
{
char ch = x.charAt(i);
char t = 'T';
char f = 'F';
if (ch=='R' || ch=='P')
{
S7.push(f);
ftf+=f;
}
if (ch == 'Q')
{
S7.push(t);
ftf+=t;
}
else if(ch=='>')
{
if(S7.peek() == 'T' && S7.prevPeek() == 'T'){
S7.pop();
S7.pop();
S7.push(t);}
else if(S7.peek() == 'T' && S7.prevPeek() == 'F'){
S7.pop();
S7.pop();
S7.push(t);}
else if(S7.peek() == 'F' && S7.prevPeek() == 'T'){
S7.pop();
S7.pop();
S7.push(f);}
else if (S7.peek() == 'F' && S7.prevPeek() == 'F'){
S7.pop();
S7.pop();
S7.push(t);}
ftf+=ch;
}
else if(ch=='<')
{
if(S7.peek() == 'T' && S7.prevPeek() == 'T'){
S7.pop();
S7.pop();
S7.push(t);}
else if(S7.peek() == 'T' && S7.prevPeek() == 'F'){
S7.pop();
S7.pop();
S7.push(f);}
else if(S7.peek() == 'F' && S7.prevPeek() == 'T'){
S7.pop();
S7.pop();
S7.push(f);}
else if (S7.peek() == 'F' && S7.prevPeek() == 'F'){
S7.pop();
S7.pop();
S7.push(t);}
ftf+=ch;
}
else if(ch=='&')
{
if(S7.peek() == 'T' && S7.prevPeek() == 'T'){
S7.pop();
S7.pop();
S7.push(t);}
else if(S7.peek() == 'T' && S7.prevPeek() == 'F'){
S7.pop();
S7.pop();
S7.push(f);}
else if(S7.peek() == 'F' && S7.prevPeek() == 'T'){
S7.pop();
S7.pop();
S7.push(f);}
else if (S7.peek() == 'F' && S7.prevPeek() == 'F'){
S7.pop();
S7.pop();
S7.push(f);}
ftf+=ch;
}
else if(ch=='v')
{
if(S7.peek() == 'T' && S7.prevPeek() == 'T'){
S7.pop();
S7.pop();
S7.push(t);}
else if(S7.peek() == 'T' && S7.prevPeek() == 'F'){
S7.pop();
S7.pop();
S7.push(t);}
else if(S7.peek() == 'F' && S7.prevPeek() == 'T'){
S7.pop();
S7.pop();
S7.push(t);}
else if (S7.peek() == 'F' && S7.prevPeek() == 'F'){
S7.pop();
S7.pop();
S7.push(f);}
ftf+=ch;
}
else if(ch=='~')
{
if(S7.peek() == 'T'){
S7.pop();
S7.push(f);}
else{
S7.pop();
S7.push(t);}
ftf+=ch;
}
}
如果需要,我可以发布更多代码,如果不清楚,我可以解释目的。
我想做类似的事情:
System.out.println("FTF = " + S7.pop());
得到正确的值。
后缀方法的中缀
public void postFix(String x)
{
Operations newList = new Operations();
output = "";
for(int i=0; i<x.length(); i++)
{
char ch = x.charAt(i);
if(ch =='>' || ch=='<' || ch=='v' || ch=='&' || ch=='~' )
{
while(!newList.empty() && (priority(newList.peek()) >= priority(ch)))
{
output+=newList.peek();
newList.pop();
}
newList.push(ch);
}
else if(ch=='(')
{
newList.push(ch);
}
else if(ch==')')
{
while (!newList.peek().equals('('))
output+=newList.pop();
newList.pop();
}
else
output+=ch;
}
while(!newList.empty())
{
output+=newList.pop();
}
System.out.println("THE INFIX = "+x);
System.out.println("THE POSTFIX = "+output);
}