399. Evaluate Division

最新推荐文章于 2021-02-15 20:04:24 发布

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最新推荐文章于 2021-02-15 20:04:24 发布

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分类专栏： LeetCode Graph 文章标签： LeetCode Graph Evaluate Division

本文链接：https://blog.youkuaiyun.com/zshouyi/article/details/73274155

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本文介绍了一种用于解决变量间关系查询的算法实现。通过构建变量间的数学关系图，并利用深度优先搜索（DFS）策略，该算法能有效地计算出两个变量间的乘除关系值。适用于诸如a/b=2.0, b/c=3.0的情况下，求解类似a/c=?这样的问题。

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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.

需要HashMap<String, HashMap<String, Double>>来存各个元素之间的关系，之后用DFS来查找元素。当元素相等时返回1.0，用一个hashset存已经visited的路径。有一点需要注意，double不能判断null，Doublr可以，因为这道题可能出现各种结果，所以判断不符合条件是最好使用null。代码如下：

public class Solution {
    public double[] calcEquation(String[][] equations, double[] values, String[][] queries) {
        HashMap<String, HashMap<String, Double>> map = new HashMap<String, HashMap<String, Double>>();
        for (int i = 0; i < equations.length; i ++) {
            String x = equations[i][0];
            String y = equations[i][1];
            if (!map.containsKey(x)) {
                map.put(x, new HashMap<String, Double>());
            }
            map.get(x).put(y, values[i]);
            if (!map.containsKey(y)) {
                map.put(y, new HashMap<String, Double>());
            }
            map.get(y).put(x, 1 / values[i]);
        }
        double[] res = new double[queries.length];
        for (int i = 0; i < res.length; i ++) {
            Double result = DFS(map, queries[i][0], queries[i][1], new HashSet<String>());
            res[i] = result == null? -1.0: result;
        }
        return res;
    }
    
    private Double DFS(HashMap<String, HashMap<String, Double>> map, String x, String y, HashSet<String> visited) {
        if (!map.containsKey(x) || !map.containsKey(y)) {
            return null;
        }
        String dupeKey = x + ":" + y;
        if (visited.contains(dupeKey)) {
            return null;
        }
        if (x.equals(y)) {
            return 1.0;
        }
        visited.add(dupeKey);
        for (String variable: map.get(x).keySet()) {
            Double result = DFS(map, variable, y, visited);
            if (result != null) {
                return result * map.get(x).get(variable);
            }
        }
        visited.remove(dupeKey);
        return null;
    }
}