Leetcode: 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.

 1 public class Solution {
 2     public double[] calcEquation(String[][] equations, double[] values, String[][] queries) {
 3         double[] res = new double[queries.length];
 4         HashMap<String, HashMap<String, Double>> map = new HashMap<String, HashMap<String, Double>>();
 5         for (int i=0; i<equations.length; i++) {
 6             String[] equation = equations[i];
 7             double value = values[i];
 8             if (!map.containsKey(equation[0])) {
 9                 map.put(equation[0], new HashMap<String, Double>());
10                 map.get(equation[0]).put(equation[0], 1.0);
11             }
12             if (!map.containsKey(equation[1])) {
13                 map.put(equation[1], new HashMap<String, Double>());
14                 map.get(equation[1]).put(equation[1], 1.0);
15             }
16             map.get(equation[0]).put(equation[1], value);
17             map.get(equation[1]).put(equation[0], 1/value);
18         }
19 
20         for (int j=0; j<queries.length; j++) {
21             res[j] = -1.0; //initialize
22             dfs(map, queries[j][0], queries[j][1], new HashSet<String>(), res, j, 1.0);
23         }
24         return res;
25     }
26     
27     public void dfs(HashMap<String, HashMap<String, Double>> map, String src, String dest, HashSet<String> visited, double[] res, int index, double pathVal) {
28         if (!map.containsKey(src) || !map.containsKey(dest)) {
29             res[index] = -1.0;
30             return;
31         }
32         if (visited.contains(src)) return;
33         HashMap<String, Double> srcNb = map.get(src);
34         if (srcNb.containsKey(dest)) {
35             res[index] = pathVal * srcNb.get(dest);
36             return;
37         }
38         visited.add(src);
39         for (Map.Entry<String, Double> entry : srcNb.entrySet()) {
40             String neibor = entry.getKey();
41             dfs(map, neibor, dest, visited, res, index, pathVal*entry.getValue());
42         }
43         visited.remove(src);
44     }
45 }