LeetCode 399. Evaluate Division--Python-DFS解法

题目地址：Evaluate Division - LeetCode

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"] ].

这道题目很好理解，就是有向图的一个权值问题。

解法是从起点开始，使用DFS,或者BFS遍历图，思路跟这道题目很像：LeetCode 207. Course Schedule–有向图找环–面试算法题–DFS递归，拓扑排序迭代–Python

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] = velse:graph[x] = {y: v}if y in graph:graph[y][x] = 1/velse:graph[y] = {x: 1/v}def dfs(s, t) -> int:if s not in graph:return -1if t == s:return 1for node in graph[s].keys():if node == t:return graph[s][node]elif node not in visited:visited.add(node)  # 添加到已访问避免重复遍历v = dfs(node, t)if v != -1:return graph[s][node]*vreturn -1res = []for qs, qt in queries:visited = set()res.append(dfs(qs, qt))return res

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