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"] ].
分析：
这是一道典型的graph题，因为从一个变量可以到另一个变量，并得到它们的关系。
1 public class Solution { 2 public double[] calcEquation(String[][] equations, double[] values, String[][] queries) { 3 Map<String, Map<String, Double>> graph = buildGraph(equations, values); 4 double[] result = new double[queries.length]; 5 6 for (int i = 0; i < queries.length; i++) { 7 result[i] = getPathWeight(queries[i][0], queries[i][1], new HashSet<>(), graph); 8 } 9 return result; 10 } 11 12 private double getPathWeight(String start, String end, Set<String> visited, 13 Map<String, Map<String, Double>> graph) { 14 if (!graph.containsKey(start)) { 15 return -1.0; 16 } 17 18 if (graph.get(start).containsKey(end)) { 19 return graph.get(start).get(end); 20 } 21 22 visited.add(start); 23 for (Map.Entry<String, Double> neighbour : graph.get(start).entrySet()) { 24 if (!visited.contains(neighbour.getKey())) { 25 double productWeight = getPathWeight(neighbour.getKey(), end, visited, graph); 26 if (productWeight != -1.0) { 27 return neighbour.getValue() * productWeight; 28 } 29 } 30 } 31 return -1.0; 32 } 33 34 private Map<String, Map<String, Double>> buildGraph(String[][] equations, double[] values) { 35 Map<String, Map<String, Double>> graph = new HashMap<>(); 36 String u, v; 37 38 for (int i = 0; i < equations.length; i++) { 39 u = equations[i][0]; 40 v = equations[i][1]; 41 graph.putIfAbsent(u, new HashMap<>()); 42 graph.get(u).put(v, values[i]); 43 graph.putIfAbsent(v, new HashMap<>()); 44 graph.get(v).put(u, 1 / values[i]); 45 } 46 return graph; 47 } 48 }
原文：https://www.cnblogs.com/beiyeqingteng/p/11287706.html