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.
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Approach #1 DFS
Binary relationship is represented as a graph usually. Does the direction of an edge matters? -- Yes. Take a / b = 2 for example, it indicates a --2--> b as well as b --1/2--> a. Thus, it is a directed weighted graph.
classSolution {publicdouble[] calcEquation(List<List<String>> equations,double[] values,List<List<String>> queries) {Map<String,Map<String,Double>> graph =buildGraph(equations, values);double[] result =newdouble[queries.length];for (int i =0; i <queries.length; i++) { result[i] =getPathWeight(queries[i][0], queries[i][1],newHashSet<>(), graph); }return result; }privatedoublegetPathWeight(String start,String end,Set<String> visited,Map<String,Map<String,Double>> graph) {if (!graph.containsKey(start)) return-1.0;if (graph.get(start).containsKey(end)) returngraph.get(start).get(end);visited.add(start);for (Map.Entry<String,Double> neighbor:graph.get(start).entrySet()) {if (!visited.contains(neighbour.getKey())) {double productWeight =getPathWeight(neighbour.getKey(), end, visited, graph);if (productWeight !=-1.0) {returnneighbour.getValue() * productWeight; } } }return-1.0; }privateMap<String,Map<String,Double>> buildGraph(String[][] equations,double[] values) {Map<String,Map<String,Double>> graph =newHashMap<>();String u, v;for (int i =0; i <equations.length; i++) { u = equations[i][0]; v = equations[i][1];graph.putIfAbsent(u,newHashMap<>());graph.get(u).put(v, values[i]);graph.putIfAbsent(v,newHashMap<>());graph.get(v).put(u,1/ values[i]); }return graph; }}