LeetCode Graph：M399. 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.

package com.graph;import java.util.*;public class Solution {    double[] res;    Map<String, Map<String, Double>> map;    Map<String, Integer> visited;       int find;    double value;    public double[] calcEquation(String[][] equations, double[] values, String[][] queries) {        //special case        if(queries==null || queries.length==0 || equations==null || equations.length==0)            return new double[0];                //normal case        //build the graph        int m = equations.length;        map = new HashMap<String, Map<String, Double>>();        //Set<String> set = new HashSet<String>();        for(int i=0; i<m; i++){            String a = equations[i][0];            String b = equations[i][1];            if(!map.containsKey(a)){                map.put(a, new HashMap<String, Double>());            }            map.get(a).put(b, values[i]);            if(!map.containsKey(b)){                map.put(b, new HashMap<String, Double>());            }            map.get(b).put(a, 1.0/values[i]);                        //get unique varaible                    }                //for each query        int n = queries.length;        res = new double[n];        for(int i=0; i<n; i++){            //dfs            value = 1.0;            find = 0;            visited = initVisited(equations);                        String cur = queries[i][0];            String tar = queries[i][1];                        //容易忽略这一步            if(!visited.containsKey(cur) || !visited.containsKey(tar)) {            res[i] = -1.0;            continue;            }                        visited.put(cur, 1);            dfs(cur, tar);            if(find==1)                res[i] = value;            else                res[i] = -1.0;        }                   return res;    }    public Map<String, Integer> initVisited(String[][] equations) {    Map<String, Integer> tvisit = new HashMap<String, Integer>();    for(int i=0; i<equations.length; i++) {    tvisit.put(equations[i][0], 0);    tvisit.put(equations[i][1], 0);    }    return tvisit;    }    public void dfs(String cur, String tar){        if(cur.equals(tar)){            find=1;            return;        }                Map<String, Double> tmap = map.get(cur);        for(Map.Entry<String, Double> tmp:tmap.entrySet()){            String next = tmp.getKey();            if(visited.get(next)==0){                visited.put(next,1);                value*=tmp.getValue();                dfs(next, tar);                if(find==1) return;//容易忽略这一步                value/=tmp.getValue();                visited.put(next,0);            }        }    }        public static void main(String[] args) {    Solution s = new Solution();    s.calcEquation(new String[][] {{"a","b"},{"b","c"}}, new double[] {2.0,3.0}, new String[][] {{"a","c"},{"b","c"},{"a","e"},{"a","a"},{"x","x"}});    }}