【Leetcode】399. Evaluate Division

题目链接：https://leetcode.com/contest/4/problems/evaluate-division/

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>> euqations, vector<double>& values, vector<pair<string, string>> query . where equations.size() == values.size(),the values are positive. this represents the equations.return vector<double>. . 
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.

思路：

用HashMap 存储图的邻接表，并用创建图节点的visited标记。这里是图节点value是string，用数组表示visited不合适，故用hashmap<string,boolean>

DFS

算法：

Map<String, Map<String, Double>> map = new HashMap<>();//邻接表public double[] calcEquation(String[][] equations, double[] values, String[][] query) {Set<String> set = new HashSet<String>();//记录表达式中出现的字符串for (int i = 0; i < equations.length; i++) {//建图set.add(equations[i][0]);set.add(equations[i][1]);Map<String, Double> m;if (map.containsKey(equations[i][0])) {m = map.get(equations[i][0]);} else {m = new HashMap<String, Double>();}m.put(equations[i][1], values[i]);map.put(equations[i][0], m);if (map.containsKey(equations[i][1])) {m = map.get(equations[i][1]);} else {m = new HashMap<String, Double>();}m.put(equations[i][0], 1.0 / values[i]);map.put(equations[i][1], m);}double result[] = new double[query.length];for (int i = 0; i < query.length; i++) {//初始化visited标记Iterator<String> it = set.iterator();Map<String, Boolean> visited = new HashMap<String, Boolean>();while (it.hasNext()) {visited.put(it.next(), false);}if (query[i][0].equals(query[i][1]) && set.contains(query[i][0])) {result[i] = 1;continue;}//dfsdouble res = dfs(query[i][0], query[i][1], 1, visited);result[i] = res;}return result;}public double dfs(String s, String t, double res, Map<String, Boolean> visited) {if (map.containsKey(s) && !visited.get(s)) {visited.put(s, true);Map<String, Double> m = map.get(s);if (m.containsKey(t)) {return res * m.get(t);} else {Iterator<String> keys = m.keySet().iterator();while (keys.hasNext()) {String key = keys.next();double state = dfs(key, t, res * m.get(key), visited);if (state != -1.0) {return state;}}}} else {return -1.0;}return -1.0;}