399. Evaluate Division

Equations are given in the format A / B = k, where A 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.

解题要点：

向大神学习！！把所有可能存在的运算方式和结果放到一个dict的dict里面。最后直接找就好了。

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]
        """
        mynbDict = collections.defaultdict(dict)
        for (first, second), val in zip(equations, values):
            mynbDict[first][first] = mynbDict[second][second] = 1.0
            mynbDict[first][second] = val
            mynbDict[second][first] = 1 / val

        for k, i, j in itertools.permutations(mynbDict, 3):
            if k in mynbDict[i] and j in mynbDict[k]:
                mynbDict[i][j] = mynbDict[i][k] * mynbDict[k][j]

        return [mynbDict[a].get(b, -1.0) for a, b in queries]