前缀时间和空间复杂度的中缀

Question

我一直在编写一个 Java 程序，使用操作数堆栈和运算符堆栈从中缀表示法转换为前缀表示法。我已经根据此处最佳答案中的伪代码实现了一个工作转换器：

conversion from infix to prefix

但是，我现在正在尝试计算上述算法的时间和空间复杂度。

我认为空间复杂度必须是 O(n)，因为我们只有两个堆栈来存储它们之间共享的输入。

考虑到时间复杂度，我不完全确定，是不是因为必须将每个子部分从中缀转换为前缀，所以是 O(n^2)？我不太确定这部分。

基本上我的问题是：我的空间复杂度结果是否正确，算法的时间复杂度是多少？

非常感谢！

编辑： 这是算法的伪代码：

Algorithm ConvertInfixtoPrefix

Purpose: Convert and infix expression into a prefix expression. Begin 
// Create operand and operator stacks as empty stacks. 
Create OperandStack
Create OperatorStack

// While input expression still remains, read and process the next token.

while( not an empty input expression ) read next token from the input expression

// Test if token is an operand or operator 
if ( token is an operand ) 
// Push operand onto the operand stack. 
    OperandStack.Push (token)
endif

// If it is a left parentheses or operator of higher precedence than the last, or the stack is empty,
else if ( token is '(' or OperatorStack.IsEmpty() or OperatorHierarchy(token) > OperatorHierarchy(OperatorStack.Top()) )
// push it to the operator stack
    OperatorStack.Push ( token )
endif

else if( token is ')' ) 
// Continue to pop operator and operand stacks, building 
// prefix expressions until left parentheses is found. 
// Each prefix expression is push back onto the operand 
// stack as either a left or right operand for the next operator. 
    while( OperatorStack.Top() not equal '(' ) 
        OperatorStack.Pop(operator) 
        OperandStack.Pop(RightOperand) 
        OperandStack.Pop(LeftOperand) 
        operand = operator + LeftOperand + RightOperand 
        OperandStack.Push(operand) 
    endwhile

// Pop the left parthenses from the operator stack. 
OperatorStack.Pop(operator)
endif

else if( operator hierarchy of token is less than or equal to hierarchy of top of the    operator stack )
// Continue to pop operator and operand stack, building prefix 
// expressions until the stack is empty or until an operator at 
// the top of the operator stack has a lower hierarchy than that 
// of the token. 
    while( !OperatorStack.IsEmpty() and OperatorHierarchy(token) lessThen Or Equal to OperatorHierarchy(OperatorStack.Top()) ) 
        OperatorStack.Pop(operator) 
        OperandStack.Pop(RightOperand) 
        OperandStack.Pop(LeftOperand) 
        operand = operator + LeftOperand + RightOperand 
        OperandStack.Push(operand)
    endwhile 
    // Push the lower precedence operator onto the stack 
    OperatorStack.Push(token)
endif
endwhile 
// If the stack is not empty, continue to pop operator and operand stacks building 
// prefix expressions until the operator stack is empty. 
while( !OperatorStack.IsEmpty() ) OperatorStack.Pop(operator) 
OperandStack.Pop(RightOperand) 
OperandStack.Pop(LeftOperand) 
operand = operator + LeftOperand + RightOperand

OperandStack.Push(operand) 
endwhile

 // Save the prefix expression at the top of the operand stack followed by popping // the      operand stack.

print OperandStack.Top()

OperandStack.Pop()

End

Answer 1

是的，O(n^2) 看起来是正确的 - 因为本质上你有一个外部和一个内部 while 循环。

编辑：O(m *n) 其中 m