【leetcode】399. Evaluate Division

题目如下：

Equations are given in the format A / B = k, whereA and B are variables represented as strings, and k is a real number (floating point number). Given some queries, return the answers. If the answer does not exist, return -1.0.

Example:
Given a / b = 2.0, b / c = 3.0. 
queries are: a / c = ?, b / a = ?, a / e = ?, a / a = ?, x / x = ? . 
return [6.0, 0.5, -1.0, 1.0, -1.0 ].

The input is: vector<pair<string, string>> equations, vector<double>& values, vector<pair<string, string>> queries , where equations.size() == values.size(), and the values are positive. This represents the equations. Return vector<double>.

According to the example above:

equations = [ ["a", "b"], ["b", "c"] ],
values = [2.0, 3.0],
queries = [ ["a", "c"], ["b", "a"], ["a", "e"], ["a", "a"], ["x", "x"] ]. 

The input is always valid. You may assume that evaluating the queries will result in no division by zero and there is no contradiction.

解题思路：我的方法是先计算equations中所有的表达式，算出所有能间接得到的值存入字典中。最后判断queries中的值是否存在于字典中即可。例如[‘a’,’b’]和[‘a’,’c’]两个组合，两者相除即可得到[‘c’,’b’]的组合。

代码如下：

class Solution(object):
    def calcEquation(self, equations, values, queries):
        """
        :type equations: List[List[str]]
        :type values: List[float]
        :type queries: List[List[str]]
        :rtype: List[float]
        """
        dic_char = {}
        dic = {}
        queue = []
        for inx,(x,y) in enumerate(equations):
            queue.append((x,y))
            dic[(x,y)] = values[inx]        while len(queue) > 0:
            x,y = queue.pop(0)
            for inx, (m,n) in enumerate(equations):
                dic_char[m] = 1
                dic_char[n] = 1
                dic[(m, n)] = values[inx]
                if (x == m and y == n) or (x == n and y == m):
                    continue
                elif x == m:
                    if (y,n) not in dic and (n,y) not in dic:
                        dic[(n,y)] = dic[(x,y)] / values[inx]
                        queue.append((n,y))
                elif x == n:
                    if (m,y) not in dic or (y,m) not in dic:
                        dic[(m,y)] = dic[(x,y)] * values[inx]
                        queue.append((m,y))
                elif y == m :
                    if (x,n) not in dic and (n,x) not in dic:
                        dic[(x,n)] = dic[(x,y)] * values[inx]
                        queue.append((x,n))
                elif y == n:
                    if (x,m) not in dic and (m,x) not in dic:
                        dic[(x,m)] = dic[(x,y)] / values[inx]
                        queue.append((x,m))
        #print dic
        #print dic_char        res = []
        for (x,y) in queries:
            if x not in dic_char or y not in dic_char:
                res.append(-1.0)
            elif x == y:
                res.append(1)
            elif (x,y) in dic:
                res.append(dic[(x,y)])
            elif (y,x) in dic:
                res.append(float(1.0)/float(dic[(y,x)]))
            else:
                res.append(-1.0)
        return res