给出方程式 A / B = k
, 其中 A
和 B
均为代表字符串的变量, k
是一个浮点型数字。根据已知方程式求解问题,并返回计算结果。如果结果不存在,则返回 -1.0
。
示例 :
给定 a / b = 2.0, b / c = 3.0
问题: a / c = ?, b / a = ?, a / e = ?, a / a = ?, x / x = ?
返回 [6.0, 0.5, -1.0, 1.0, -1.0 ]
输入为: vector<pair<string, string>> equations, vector<double>& values, vector<pair<string, string>> queries
(方程式,方程式结果,问题方程式), 其中 equations.size() == values.size()
,即方程式的长度与方程式结果长度相等(程式与结果一一对应),并且结果值均为正数。以上为方程式的描述。 返回vector<double>
类型。
基于上述例子,输入如下:
equations(方程式) = [ ["a", "b"], ["b", "c"] ], values(方程式结果) = [2.0, 3.0], queries(问题方程式) = [ ["a", "c"], ["b", "a"], ["a", "e"], ["a", "a"], ["x", "x"] ].
输入总是有效的。你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。
思路:
- 将每个字符串变量以有向图的形式构建,权值即为商,问题方程式就归结为起点与终点是否有通路
#include <string>
#include <vector>
#include <set>
#include <map>
using namespace std;
class Solution {
public:
vector<double> calcEquation(vector<pair<string, string>> equations, vector<double>& values, vector<pair<string, string>> queries) {
set<string> points;
map<pair<string, string>, double> mapKeyValues;
for (int i = 0; i < equations.size(); ++i)
{
points.insert(equations[i].first);
points.insert(equations[i].second);
mapKeyValues[equations[i]] = values[i];
mapKeyValues[{equations[i].second, equations[i].first}] = 1 / values[i];
}
vector<double> result;
for (int i = 0; i < queries.size(); ++i)
{
set<string> visits;
visits.insert(queries[i].first);
result.push_back(dfs(queries[i], queries[i].first, points, visits, mapKeyValues));
mapKeyValues[queries[i]] = result[result.size() - 1];
}
return result;
}
double dfs(pair<string, string>& query, string nowPosition, set<string>& points, set<string>& visits, map<pair<string, string>, double>& mapKeyValues)
{
if (points.find(query.first) == points.end() || points.find(query.second) == points.end())
return -1.0;
else if (query.first == query.second || query.second == nowPosition)
return 1.0;
for (string nextPoint : points)
{
if (mapKeyValues.find({ nowPosition , nextPoint}) != mapKeyValues.end() && visits.find(nextPoint) == visits.end())
{
visits.insert(nextPoint);
double temp = dfs(query, nextPoint, points, visits, mapKeyValues);
if (temp == -1.0)
visits.erase(nextPoint);
else
return mapKeyValues[{nowPosition, nextPoint}] * temp;
}
}
return -1.0;
}
};