Leetcode 399 Evaluate Division
題目原文
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.
題意分析
給出多個觸發等式作爲已知條件，判斷是否能得到目標計算的結果，如果能，返回結果，不能，返回-1.0.對於分子分母相等的目標計算式，如果運算元素在已知條件等式中，則返回結果1，不然結果仍爲-1.0.分子分母都是由string表示的一個數。
解法分析
當目標計算式就在已知條件中時可以直接得到結果，如果不在其中，需要判斷經過多次相乘，能否得到最終結果，由於a/b=2.0，同時可得b/a=0.5,加入目標爲a/c，則該問題可以看做a和c之間的尋路問題，a->b的權重爲2.0,b->a的權重爲0.5.因此可以利用已知等式構造一個圖，從起點a開始尋路到c，利用深度優先搜索，遍歷a的相鄰點，並延某一點一直往下走，注意走過的點要用attended進行記錄。
由本題可以看到，BFS和DFS的運用最主要是將問題轉化爲一個圖問題，樹的遍歷只是一個特例。C++代碼如下：
class Solution {
private:
double value;
string b;
unordered_map<string, vector<pair<string,double>>> myGraph;
unordered_set<string> attended;
public:
double findWay(string a){
double res;
if(attended.count(a))
return -1.0;
attended.insert(a);
for(auto p:myGraph[a]){
value*=p.second;
if(p.first==b)
return value;
res=findWay(p.first);
if(res==-1.0)
value/=p.second;
else
return res;
}
return -1.0;
}
vector<double> calcEquation(vector<pair<string, string>> equations, vector<double>& values, vector<pair<string, string>> queries){
for(int i=0;i<values.size();i++){
myGraph[equations[i].first].push_back(make_pair(equations[i].second,values[i]));
myGraph[equations[i].second].push_back(make_pair(equations[i].first,1.0/values[i]));//make the graph
}
vector<double> res;
for(auto q:queries){
b=q.second;
value=1;
attended.clear();
if(myGraph.count(q.first)==0||myGraph.count(q.second)==0)
res.push_back(-1.0);
else{
if(q.first==q.second)
res.push_back(1.0);
else
res.push_back(findWay(q.first));
}
}
return res;
}
};
