#Leet Code# Evaluate Reverse Polish Notation

描述：计算逆波兰表达法的结果

Sample：

  ["", "", "+", "", "*"] -> ((2 + 1) * 3) -> 9
  ["", "", "", "/", "+"] -> (4 + (13 / 5)) -> 6

使用stack实现：

 def is_op(c):
     return c in ['+', '-', '*', '/']
 
 def divide(x, y):
     if (x * y) < 0:
         return -1 * ((-x)/y)
     return x/y
 
 class Solution:
     # @param tokens, a list of string
     # @return an integer
     def evalRPN(self, tokens):
         opDict = {'+': lambda x,y: x+y,
                   '-': lambda x,y: x-y,
                   '*': lambda x,y: x*y,
                   '/': divide}
         record = []
 
         for item in tokens:
             if is_op(item):
                 second = record.pop()
                 first = record.pop()
                 record.append(opDict[item](first, second))
             else:
                 record.append(int(item))
         return record[0]

使用树实现：

 def is_op(c):
     return c in ['+', '-', '*', '/']
 
 def divide(x, y):
     if (x * y) < 0:
         return -1 * ((-x)/y)
     return x/y
 
 class Tree:
     def __init__(self, data):
         self.data = data
         self.parent = None
         self.left = None
         self.right = None
 
 class Solution:
     # @param tokens, a list of string
     # @return an integer
     def __init__(self):
         self.opDict = {'+': lambda x,y: x+y,
                   '-': lambda x,y: x-y,
                   '*': lambda x,y: x*y,
                   '/': divide}
 
     def builtTree(self, tokens):
         if not is_op(tokens[-1]):
             return int(tokens[-1])
 
         # if element is an operator
         cur_tree = Tree(tokens[-1])
         top_tree = cur_tree
 
         for item in tokens[-2::-1]:
             if cur_tree.right is None:
                 if is_op(item):
                     cur_tree.right = Tree(item)
                     cur_tree.right.parent = cur_tree
                     cur_tree = cur_tree.right
                 else:
                     cur_tree.right = int(item)
 
                 if cur_tree.right and cur_tree.left:
                     cur_tree = self.getUpperNode(cur_tree)
                 continue
 
             if cur_tree.left is None:
                 if is_op(item):
                     cur_tree.left = Tree(item)
                     cur_tree.left.parent = cur_tree
                     cur_tree = cur_tree.left
                 else:
                     cur_tree.left = int(item)
 
                 if cur_tree.right is not None and cur_tree.left is not None:
                     cur_tree = self.getUpperNode(cur_tree)
 
         return top_tree
 
     # Move to upper node if cur node if full. If top_node return.
     def getUpperNode(self, node):
         while node.right is not None and node.left is not None:
             if node.parent is None:
                 return node
 
             node = node.parent
 
         return node
 
     def getValue(self, node):
         if type(node) is type(1):
             return node
         else:
             return self.getResult(node)
 
     def getResult(self, treeNode):
         leftValue = self.getValue(treeNode.left)
         rightValue = self.getValue(treeNode.right)
 
         result = self.opDict[treeNode.data](leftValue, rightValue)
 
         return result
 
     def evalRPN(self, tokens):
         topNode = self.builtTree(tokens)
 
         if type(topNode) is type(1):
             return topNode
         else:
             resultNum = self.getResult(topNode)
             return resultNum

备注-1：if cur_node.right is None 不能用 if cur_node.right 因为cur_node.right 如果是数字0的话会有问题 当然 不转成int的话直接存string等到运算时再转int应该就可以这样写了

备注-2：python的除法跟c++不太一样 3/-5 = -1

结论：

根据问题的具体特性，选择合适的数据结构解决问题会差别很大