# LeetCode 399. Evaluate Division--Python-DFS解法-图

2020/02/02

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

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"] ].
``````

Python-DFS解法如下：

``````class Solution:

def calcEquation(self, equations: List[List[str]], values: List[float], queries: List[List[str]]) -> List[float]:
graph = {}
for (x, y), v in zip(equations, values):
if x in graph:
graph[x][y] = v
else:
graph[x] = {y: v}
if y in graph:
graph[y][x] = 1/v
else:
graph[y] = {x: 1/v}

def dfs(s, t) -> int:
if s not in graph:
return -1
if t == s:
return 1
for node in graph[s].keys():
if node == t:
return graph[s][node]
elif node not in visited: