Question Definition
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
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.
Java Solution
class Solution {
private Map<String, List<Object[]>> graph;
private Set<String> visited;
public double[] calcEquation(String[][] equations, double[] values, String[][] queries) {
graph = new HashMap<>();
visited = new HashSet<>();
for(int i = 0; i < equations.length; i++) {
String[] equation = equations[i];
graph.putIfAbsent(equation[0], new LinkedList<>());
graph.putIfAbsent(equation[1], new LinkedList<>());
graph.get(equation[0]).add(new Object[]{equation[1], values[i]});
graph.get(equation[1]).add(new Object[]{equation[0], 1.0 / values[i]});
}
double[] result = new double[queries.length];
for(int i = 0; i < queries.length; i++){
result[i] = gfs(queries[i][0], queries[i][1]);
visited.clear();
}
return result;
}
private double gfs(String start, String end){
if(!graph.containsKey(start) || !graph.containsKey(end))
return -1;
if(start.equals(end)) return 1.0;
for(int i = 0; i < graph.get(start).size(); i++){
String edge = (String)graph.get(start).get(i)[0];
if(edge.equals(end)) return (double)graph.get(start).get(i)[1];
if(!visited.contains(edge)){
visited.add(edge);
double temp = gfs(edge, end);
if(temp != -1)
return (double)graph.get(start).get(i)[1] * temp;
}
}
return -1;
}
}
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