图 floyd

399. Evaluate Division Medium

You are given an array of variable pairs equations and an array of real numbers values, where equations[i] = [Ai, Bi] and values[i] represent the equation Ai / Bi = values[i]. Each Ai or Bi is a string that represents a single variable.

You are also given some queries, where queries[j] = [Cj, Dj] represents the jth query where you must find the answer for Cj / Dj = ?.

Return the answers to all queries. If a single answer cannot be determined, return -1.0.

Note: The input is always valid. You may assume that evaluating the queries will not result in division by zero and that there is no contradiction.

Example 1:

Input: equations = [["a","b"],["b","c"]], values = [2.0,3.0], queries = [["a","c"],["b","a"],["a","e"],["a","a"],["x","x"]]
Output: [6.00000,0.50000,-1.00000,1.00000,-1.00000]
Explanation: 
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 ]

Example 2:

Input: equations = [["a","b"],["b","c"],["bc","cd"]], values = [1.5,2.5,5.0], queries = [["a","c"],["c","b"],["bc","cd"],["cd","bc"]]
Output: [3.75000,0.40000,5.00000,0.20000]

Example 3:

Input: equations = [["a","b"]], values = [0.5], queries = [["a","b"],["b","a"],["a","c"],["x","y"]]
Output: [0.50000,2.00000,-1.00000,-1.00000] 

Constraints:

• 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 consist of lower case English letters and digits.

class Solution {
    public double[] calcEquation(List<List<String>> equations, double[] values, List<List<String>> queries) {
        Map<String,Map<String,Double>> map = new HashMap();
        
        for(int i=0;i<values.length;i++){
            String first = equations.get(i).get(0),second = equations.get(i).get(1);
            map.put(first,map.getOrDefault(first,new HashMap()));
            map.put(second,map.getOrDefault(second,new HashMap()));
            Map<String,Double> firstMap = map.get(first);
            Map<String,Double> secondMap = map.get(second);
            firstMap.put(first,1d);
            secondMap.put(second,1d);
            firstMap.put(second,values[i]);
            secondMap.put(first,1/values[i]);
        }
        for(String key:map.keySet()){
            Map<String,Double> currmap = map.get(key);
            for(String key1:currmap.keySet()){
                for(String key2:currmap.keySet()){
                    if(key.equals(key1) || key.equals(key2) || key1.equals(key2)) continue;
                    double value12 = map.get(key1).get(key)/map.get(key2).get(key);
                    map.get(key1).put(key2,value12);
                    map.get(key2).put(key1,1/value12);
                }
            }
        }
        double[] results = new double[queries.size()];
        for(int i=0;i<queries.size();i++){
            String first = queries.get(i).get(0),second = queries.get(i).get(1);
            if(map.get(first)!=null && map.get(first).get(second)!=null)
                results[i]=map.get(first).get(second);
            else
                results[i]=-1;
        }
        return results;
    }
}

 

标签：map,get,floyd,values,key,queries,equations
来源： https://www.cnblogs.com/cynrjy/p/15852491.html