399. Evaluate Division

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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.

DFS:

public class Solution {    public double[] calcEquation(String[][] equations, double[] values, String[][] queries) {        Map<String,Map<String,Double>> map=new HashMap<>();// 邻接表        Set<String> set=new HashSet<>();        for(int i=0;i<equations.length;i++){    //创建图            set.add(equations[i][0]);            set.add(equations[i][1]);            Map<String,Double> m;            if(map.containsKey(equations[i][0])){                m=map.get(equations[i][0]);            } else{                m=new HashMap<>();            }            m.put(equations[i][1],values[i]);            map.put(equations[i][0],m);            if(map.containsKey(equations[i][1])){                m=map.get(equations[i][1]);            }else{                m=new HashMap<>();            }            m.put(equations[i][0],1.0/values[i]);            map.put(equations[i][1],m);        }        double[] res=new double[queries.length];        for(int i=0;i<queries.length;i++){            Map<String,Boolean> visited=new HashMap<>();   //用于判断是否访问过节点            for(String it:set){                   visited.put(it,false);            }            if(queries[i][0].equals(queries[i][1])&&set.contains(queries[i][0])){                res[i]=1;                continue;            }            double result=dfs(map,queries[i][0],queries[i][1],1.0,visited);            res[i]=result;        }        return res;    }    public double dfs(Map<String,Map<String,Double>> map,String t,String s,double res,Map<String,Boolean> visited){        if(map.containsKey(t)&&!visited.get(t)){            visited.put(t,true);            Map<String,Double> m=map.get(t);            if(m.containsKey(s))                return res*m.get(s);            else{                for(String it:m.keySet()){                    double tmp=dfs(map,it,s,res*m.get(it),visited);                    if(tmp!=-1.0){                        return tmp;                    }                }            }        }else{            return -1.0;        }        return -1.0;    }}

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