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.

Solution

import java.lang.Double;

public class Solution {
    public double[] calcEquation(String[][] equations, double[] values, String[][] queries) {
        if(equations == null || values == null || queries == null) return null;
        Map<String, Map<String, Double> > map = new HashMap<String, Map<String, Double>>();
        for(int i = 0; i < equations.length; i++) {
            if(map.get(equations[i][0]) == null) map.put(equations[i][0], new HashMap<String, Double>());
            if(map.get(equations[i][1]) == null) map.put(equations[i][1], new HashMap<String, Double>());

            map.get(equations[i][0]).put(equations[i][1], values[i]);
            map.get(equations[i][1]).put(equations[i][0], 1.0/values[i]);
        }

        double ans[] = new double[queries.length];
        for(int i = 0; i < queries.length; i++) {
            if( map.get(queries[i][0]) == null || map.get(queries[i][1]) == null) {
                ans[i] = -1;
            } else {
                if(queries[i][0].equals(queries[i][1])) {
                    ans[i] = 1.0;
                    continue;
                }

                Double a = dfs(new ArrayList<String>(), 1.0, queries[i][0], queries[i][1], map);
                ans[i] = a == null ? -1 : a;
            }
        }
        return ans;
    }

    public Double dfs(List<String> path, double val, String cur, String target, Map<String, Map<String, Double>> map) {

        for(String key: map.get(cur).keySet()) {
            if(path.contains(key)) continue;

            double v = val * map.get(cur).get(key);
            if(key.equals(target)) return v;

            path.add(cur);
            Double aa = dfs(path, v, key, target, map);
            path.remove(path.size() - 1);

            if(aa != null) return aa;
        }
        return null;
    }

}