题目
给你一个变量对数组 equations 和一个实数值数组 values 作为已知条件，其中 equations[i] = [Ai, Bi] 和 values[i] 共同表示等式 Ai / Bi = values[i] 。每个 Ai 或 Bi 是一个表示单个变量的字符串。
另有一些以数组 queries 表示的问题，其中 queries[j] = [Cj, Dj] 表示第 j 个问题，请你根据已知条件找出 Cj / Dj = ? 的结果作为答案。
返回 所有问题的答案 。如果存在某个无法确定的答案，则用 -1.0 替代这个答案。如果问题中出现了给定的已知条件中没有出现的字符串，也需要用 -1.0 替代这个答案。
注意：输入总是有效的。你可以假设除法运算中不会出现除数为 0 的情况，且不存在任何矛盾的结果。
注意：未在等式列表中出现的变量是未定义的，因此无法确定它们的答案。
示例 1：
输入：equations = [["a","b"],["b","c"]], values = [2.0,3.0], queries = [["a","c"],["b","a"],["a","e"],["a","a"],["x","x"]]
输出：[6.00000,0.50000,-1.00000,1.00000,-1.00000]
解释：
条件：a / b = 2.0, b / c = 3.0
问题：a / c = ?, b / a = ?, a / e = ?, a / a = ?, x / x = ?
结果：[6.0, 0.5, -1.0, 1.0, -1.0 ]
注意：x 是未定义的 => -1.0
示例 2：
输入：equations = [["a","b"],["b","c"],["bc","cd"]], values = [1.5,2.5,5.0], queries = [["a","c"],["c","b"],["bc","cd"],["cd","bc"]]
输出：[3.75000,0.40000,5.00000,0.20000]
示例 3：
输入：equations = [["a","b"]], values = [0.5], queries = [["a","b"],["b","a"],["a","c"],["x","y"]]
输出：[0.50000,2.00000,-1.00000,-1.00000]
提示：
1 <= equations.length <= 20
equations[i].length == 2
1 <= Ai.length, Bi.length <= 5
values.length == equations.length
0.0 < values[i] <= 20.0
1 <= queries.length <= 20
queries[i].length == 2
1 <= Cj.length, Dj.length <= 5
Ai, Bi, Cj, Dj 由小写英文字母与数字组成
思路
一个带权值的无向图。就是求给两个节点，看能否在图中由一个节点遍历到另一个节点，可以的话边的权值的乘积就是答案。
代码
java
class Solution {
public class Graph{
public HashMap<String,Node> nodes;
public HashSet<Edge> edges;
public Graph(){
nodes = new HashMap<>();
edges = new HashSet<>();
}
}
public class Node{
public String value;
public int in;
public int out;
public ArrayList<Node> nexts;
public ArrayList<Edge> edges;
public Node(String value){
this.value = value;
in = 0;
out = 0;
nexts = new ArrayList<>();
edges = new ArrayList<>();
}
}
public class Edge{
public double weight;
public Node from;
public Node to;
public Edge(double weight, Node from, Node to){
this.weight = weight;
this.from = from;
this.to = to;
}
}
public double[] calcEquation(List<List<String>> equations, double[] values, List<List<String>> queries) {
ArrayList<Double> ans = new ArrayList<>();
//1.构建双向图
Graph graph = createGraph(equations,values);
// HashMap<String,Node> nodes= graph.nodes;
//2.遍历双向图的节点，找是否出现变量
for(int i=0;i<queries.size();i++){
String startVal = queries.get(i).get(0);
String endVal = queries.get(i).get(1);
double distance = 0;
if(graph.nodes.containsKey(startVal)&&graph.nodes.containsKey(endVal)){
Node fromNode = graph.nodes.get(startVal);
Node toNode = graph.nodes.get(endVal);
HashSet<Node> vis = new HashSet<>();
ans.add(bfs(fromNode,toNode,vis));
}else{
ans.add(-1.0);
}
}
double[] finall = new double[ans.size()];
for(int i=0;i<ans.size();i++){
finall[i]=ans.get(i);
}
return finall;
}
public double bfs(Node fromNode,Node toNode,HashSet<Node> vis){
if(fromNode==toNode) return 1.0;
vis.add(fromNode);
for(Edge nextEdge: fromNode.edges){
Node nextNode = nextEdge.to;
if(!vis.contains(nextNode)){
double distance = nextEdge.weight;
double res = bfs(nextNode,toNode,vis);
if(res!=-1.0) return distance*res;
}
}
return -1.0;
}
public Graph createGraph(List<List<String>> equations,double[] values){
Graph graph = new Graph();
for(int i=0;i<equations.size();i++){
String fromVal = equations.get(i).get(0);
String toVal = equations.get(i).get(1);
if(!graph.nodes.containsKey(fromVal))graph.nodes.put(fromVal, new Node(fromVal));
if(!graph.nodes.containsKey(toVal)) graph.nodes.put(toVal, new Node(toVal));
Node fromNode = graph.nodes.get(fromVal);
Node toNode = graph.nodes.get(toVal);
Edge edge = new Edge(values[i], fromNode, toNode);
fromNode.out++;
toNode.in++;
fromNode.nexts.add(toNode);
fromNode.edges.add(edge);
graph.edges.add(edge);
Edge edge1 = new Edge(1/values[i],toNode, fromNode);
toNode.out++;
fromNode.in++;
toNode.nexts.add(fromNode);
toNode.edges.add(edge1);
graph.edges.add(edge1);
}
return graph;
}
}执行用时分布1ms，击败81.63%使用 Java 的用户。应该不用再优化了