问题简介

给定一些由变量组成的等式组，然后根据这些等式推算出所闻的等式的结果，如果无法推算，则返回-1.0。
比如：

给定等式组
a / b = 2.0, b / c = 3.0
求出
a / c = ?, b / a = ?, a / e = ?, a / a = ?, x / x = ?
返回结果为
[ 6.0, 0.5, -1.0, 1.0, -1.0 ]

注：给定的等式组不存在结果为0的情况。

解题思路

先构造一个数据结构，然后根据给定的等式构造一个map，便于访问，这其实就相当与是一幅稀疏图。
要注意的是，这里在保存$a/b=2.0$这样的等式时，同时也保存了$b/a=0.5$。

// 构造一个结构体
struct 节点{是否访问过;unordered_map<除数, 结果>;
};// 构造稀疏图
unordered_map<变量, 节点>;

然后根据询问的内容做深度优先遍历即可。

源代码

struct Node{bool visited;unordered_map<string, double> children;Node(){visited = false;}
};class Solution {
private:vector<double> res;private:bool dfs(unordered_map<string, Node*>& mp, pair<string, string>& p, double r){if (mp.find(p.first) != mp.end() && !mp[p.first]->visited){mp[p.first]->visited = true;if (mp[p.first]->children.find(p.second) != mp[p.first]->children.end()){res.push_back(r * mp[p.first]->children[p.second]);mp[p.first]->visited = false;return true;}else{for (auto iter = mp[p.first]->children.begin(); iter != mp[p.first]->children.end(); iter++){pair<string, string> new_p = make_pair(iter->first, p.second);if (dfs(mp, new_p, r * iter->second)){mp[p.first]->visited = false;return true;}}}mp[p.first]->visited = false;}return false;}public:vector<double> calcEquation(vector<pair<string, string>> equations, vector<double>& values, vector<pair<string, string>> queries) {// 构造图unordered_map<string, Node*> mp;for (int i = 0; i < equations.size(); i++){// 保存a/bif (mp.find(equations[i].first) == mp.end()){mp[equations[i].first] = new Node();}Node* tmp = mp[equations[i].first];if (tmp->children.find(equations[i].second) == tmp->children.end()){tmp->children[equations[i].second] = values[i];}// 保存b/aif (mp.find(equations[i].second) == mp.end()){mp[equations[i].second] = new Node();}tmp = mp[equations[i].second];if (tmp->children.find(equations[i].first) == tmp->children.end()){tmp->children[equations[i].first] = 1.0 / values[i];}}// 开始执行queryfor (int i = 0; i < queries.size(); i++){if (queries[i].first == queries[i].second){if (mp.find(queries[i].first) != mp.end())res.push_back(1.0);elseres.push_back(-1.0);}else{if (!dfs(mp, queries[i], 1.0)){res.push_back(-1.0);}}}return res;}
};