399. Evaluate Division

Posted 2021-01-15 gopanama

https://leetcode.com/problems/evaluate-division/discuss/88169/Java-AC-Solution-using-graph

建立graph dfs
value是按照值建立的 这样dfs的时候可以直接乘

 

 

 1 class Solution {
 2     public double[] calcEquation(String[][] equations, double[] values, String[][] queries) {
 3         HashMap<String, List<String>> pairs = new HashMap<>();
 4         HashMap<String, List<Double>> value = new HashMap<>();
 5         for(int i = 0; i < equations.length; i++){
 6             if(!pairs.containsKey(equations[i][0])){
 7                 pairs.put(equations[i][0], new ArrayList<>());
 8                 value.put(equations[i][0], new ArrayList<>());
 9             }
10             if(!pairs.containsKey(equations[i][1])){
11                 pairs.put(equations[i][1], new ArrayList<>());
12                 value.put(equations[i][1], new ArrayList<>());
13             }
14             pairs.get(equations[i][0]).add(equations[i][1]);
15             pairs.get(equations[i][1]).add(equations[i][0]);
16             value.get(equations[i][0]).add(values[i]);
17             value.get(equations[i][1]).add(1 / values[i]);
18         }
19         
20         double[] result = new double[queries.length];
21         int i = 0;
22         for(String[] query : queries){
23             if(!pairs.containsKey(query[0]) || !pairs.containsKey(query[1])){
24                 result[i++] = -1.0;
25                 continue;
26             }
27             double res = dfs(pairs, value, new HashSet<String>(), query[0], query[1], 1.0);
28             if(res == 0.0){
29                 result[i++] = -1.0;
30             }else{
31                 result[i++] = res;
32             }
33         }
34         return result;
35         
36     }
37     
38     public double dfs(HashMap<String, List<String>> pairs, HashMap<String, List<Double>> values, HashSet<String> set, String start, String end, double num){
39         if(start.equals(end)) return num;
40         if(set.contains(start)) return 0.0;
41         set.add(start);
42         
43         List<String> list = pairs.get(start);
44         List<Double> value = values.get(start);
45         double res = 0.0;
46         for(int i = 0; i < list.size(); i++){
47             if(!set.contains(list.get(i))){
48                 res = dfs(pairs, values, set, list.get(i), end, num * value.get(i));
49             }
50             if(res != 0.0) break;
51         }
52         return res;
53     }
54 }