LCR 111. 除法求值

LCR 111. 除法求值

给定一个变量对数组 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。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。

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 ]

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

class Solution {
    Map<String, Map<String, Double>> nodesEdges = new HashMap<String, Map<String, Double>>();

    public double[] calcEquation(List<List<String>> equations, double[] values, List<List<String>> queries) {
        int edgesCount = equations.size();
        for (int i = 0; i < edgesCount; i++) {
            List<String> equation = equations.get(i);
            double value = values[i];
            String node0 = equation.get(0), node1 = equation.get(1);
            nodesEdges.putIfAbsent(node0, new HashMap<String, Double>());
            nodesEdges.get(node0).put(node1, value);
            nodesEdges.putIfAbsent(node1, new HashMap<String, Double>());
            nodesEdges.get(node1).put(node0, 1.0 / value);
        }
        int queriesCount = queries.size();
        double[] answer = new double[queriesCount];
        for (int i = 0; i < queriesCount; i++) {
            String start = queries.get(i).get(0), end = queries.get(i).get(1);
            answer[i] = nodesEdges.containsKey(start) && nodesEdges.containsKey(end) ? bfs(start, end) : -1.0;
        }
        return answer;
    }

    public double bfs(String start, String end) {
        double queryValue = -1.0;
        Set<String> visited = new HashSet<String>();
        Queue<String> nodeQueue = new ArrayDeque<String>();
        Queue<Double> valueQueue = new ArrayDeque<Double>();
        visited.add(start);
        nodeQueue.offer(start);
        valueQueue.offer(1.0);
        while (queryValue < 0 && !nodeQueue.isEmpty()) {
            String node = nodeQueue.poll();
            double value = valueQueue.poll();
            if (node.equals(end)) {
                queryValue = value;
            } else {
                Map<String, Double> adjacent = nodesEdges.getOrDefault(node, new HashMap<String, Double>());
                Set<Map.Entry<String, Double>> entries = adjacent.entrySet();
                for (Map.Entry<String, Double> entry : entries) {
                    String nextNode = entry.getKey();
                    double nextValue = entry.getValue();
                    if (visited.add(nextNode)) {
                        nodeQueue.offer(nextNode);
                        valueQueue.offer(value * nextValue);
                    }
                }
            }
        }
        return queryValue;
    }
}