leetcode [399]Evaluate Division
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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.
题目大意:
给一系列表达式的值,查询某一表达式是否能以给定表达式的值求出。
解法:
根据表达式的值可以组成图,比如a/b=2,那么就是a到b存在边,且边上的权重为2,b到a也是存在边的,且权重为1/2.
java:
class Solution private double dfs(String start,String end,Map<String,ArrayList<String>> pairs, Map<String, ArrayList<Double>> values, HashSet<String> set, double value) if (set.contains(start)) return 0.0; if (!pairs.containsKey(start)) return 0.0; if (start.equals(end)) return value; set.add(start); ArrayList<String> strList=pairs.get(start); ArrayList<Double> valueList=values.get(start); double tmp=0.0; for (int i=0;i<strList.size();i++) tmp=dfs(strList.get(i),end,pairs,values,set,value*valueList.get(i)); if (tmp!=0) break; set.remove(start); return tmp; public double[] calcEquation(List<List<String>> equations, double[] values, List<List<String>> queries) Map<String,ArrayList<String>> pairs=new HashMap<>(); Map<String,ArrayList<Double>> valuesPair=new HashMap<>(); for (int i=0;i<equations.size();i++) pairs.computeIfAbsent(equations.get(i).get(0),k->new ArrayList<>()).add(equations.get(i).get(1)); valuesPair.computeIfAbsent(equations.get(i).get(0),k->new ArrayList<>()).add(values[i]); pairs.computeIfAbsent(equations.get(i).get(1),k->new ArrayList<>()).add(equations.get(i).get(0)); valuesPair.computeIfAbsent(equations.get(i).get(1),k->new ArrayList<>()).add(1/values[i]); double[] res=new double[queries.size()]; for (int i=0;i<queries.size();i++) res[i]=dfs(queries.get(i).get(0),queries.get(i).get(1),pairs,valuesPair,new HashSet<>(),1.0); if (res[i]==0.0) res[i]=-1.0; return res;
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