原题链接在这里：https://leetcode.com/problems/evaluate-division/description/
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.
题解：
建立graph, 和graph中的edge的weight.
在求新的query时做dfs. 求结果.
Time Complexity: O(queries.length*(V+E)). dfs 用时O(V+E). E = equations.length. V = graph.size().
Space: O(V). regardless res.
AC Java:
1 class Solution {
2 public double[] calcEquation(String[][] equations, double[] values, String[][] queries) {
3 HashMap<String, List<String>> graph = new HashMap<String, List<String>>();
4 HashMap<String, List<Double>> weight = new HashMap<String, List<Double>>();
5 for(int i = 0; i<equations.length; i++){
6 String [] equation = equations[i];
7 if(!graph.containsKey(equation[0])){
8 graph.put(equation[0], new ArrayList<String>());
9 weight.put(equation[0], new ArrayList<Double>());
10 }
11 graph.get(equation[0]).add(equation[1]);
12 weight.get(equation[0]).add(values[i]);
13
14 if(values[i] == 0.0){
15 continue;
16 }
17
18 if(!graph.containsKey(equation[1])){
19 graph.put(equation[1], new ArrayList<String>());
20 weight.put(equation[1], new ArrayList<Double>());
21 }
22 graph.get(equation[1]).add(equation[0]);
23 weight.get(equation[1]).add(1.0/values[i]);
24 }
25
26 double [] res = new double[queries.length];
27 for(int i = 0; i<queries.length; i++){
28 String [] query = queries[i];
29 res[i] = dfs(query[0], query[1], graph, weight, new HashSet<String>(), 1.0);
30 if(res[i] == 0.0){
31 res[i] = -1.0;
32 }
33 }
34
35 return res;
36 }
37
38 private double dfs(String start, String end, Map<String, List<String>> graph, Map<String, List<Double>> weight, Set<String> visited, double cur){
39 if(visited.contains(start) || !graph.containsKey(start)){
40 return 0.0;
41 }
42
43 if(start.equals(end)){
44 return cur;
45 }
46
47 double res = 0.0;
48 visited.add(start);
49 for(int i = 0; i<graph.get(start).size(); i++){
50 res = dfs(graph.get(start).get(i), end, graph, weight, visited, cur*weight.get(start).get(i));
51 if(res != 0.0){
52 break;
53 }
54 }
55 visited.remove(start);
56 return res;
57 }
58 }
原文地址：https://www.cnblogs.com/Dylan-Java-NYC/p/8452702.html