leetcode 399. Evaluate Division

题目：

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.

 

import java.util.*;

public class Solution {
    public double[] calcEquation(String[][] equations, double[] values, String[][] queries) {
        Map<String,Map<String,Double>> eqmap = new HashMap<String, Map<String, Double>>();
        double[] result = new double[queries.length];
        for(int i=0;i<equations.length;i++){
            Map<String,Double> temp = new HashMap<String, Double>();
            if(eqmap.containsKey(equations[i][0]))
                eqmap.get(equations[i][0]).put(equations[i][1],values[i]);
            else{
                temp.put(equations[i][1],values[i]);
                eqmap.put(equations[i][0],temp);
            }
            if(eqmap.containsKey(equations[i][1]))
                eqmap.get(equations[i][1]).put(equations[i][0],1/values[i]);
            else{
                temp = new HashMap<String, Double>();
                temp.put(equations[i][0],1/values[i]);
                eqmap.put(equations[i][1],temp);
            }
        }
        for(int j=0;j<queries.length;j++){
            Stack<String> strK = new Stack<String>();
            Set<String> strS = new HashSet<String>();
            result[j] = getResult(eqmap,queries[j],strK,strS);
        }
        return result;
    }
    private double getResult(Map<String,Map<String,Double>> eqmap, String[] query, Stack<String> strK,Set<String> strS){
        if(eqmap.containsKey(query[0])&&eqmap.containsKey(query[1])){
            Map<String, Double> tempmap = eqmap.get(query[0]);
            if (tempmap.containsKey(query[1])) {
                return tempmap.get(query[1]);
            }
            else{
                Iterator ite = eqmap.get(query[0]).entrySet().iterator();
                while(ite.hasNext()){
                    Map.Entry ent = (Map.Entry)ite.next();
                    if(!strS.add((String)ent.getKey()))
                        continue;
                    strK.push(query[0]);
                    double result = (Double) ent.getValue() * getResult(eqmap,new String[]{(String) ent.getKey(),query[1]},strK,strS);
                    if(result>0)
                        return result;
                    else strK.pop();
                }
            }
        }
        return -1.0;
    }
}