【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.

可以用dfs和并查集

dfs解法：

class Solution {
public:
    vector<double> calcEquation(vector<pair<string, string>> equations, vector<double>& values, vector<pair<string, string>> queries) {
        unordered_map<string,vector<pair<string,double>>> adj;
        unordered_set<string> vis;
        for(int i=0;i<equations.size();i++){
            string s1=equations[i].first,s2=equations[i].second;
            double x=values[i];
            adj[s1].push_back(make_pair(s2,x));
            adj[s2].push_back(make_pair(s1,1/x));
        }
        vector<double> ans;
        for(int i=0;i<queries.size();i++){
            string s1=queries[i].first,s2=queries[i].second;
            double val=dfs(s1,s2,adj,vis);
            ans.push_back(val);
            vis.clear();
        }
        return ans;
    }
    double dfs(string s1,string s2,unordered_map<string,vector<pair<string,double>>> &adj,unordered_set<string> &vis){
        if(adj[s1].size()==0||adj[s2].size()==0)return -1;
        if(s1.compare(s2)==0) return 1;
        if(vis.find(s1)!=vis.end()) return -1;
        vis.insert(s1);
        for(auto itr=adj[s1].begin();itr!=adj[s1].end();++itr){
            string next = itr->first;
            double num = itr->second;
            double ans = dfs(next,s2,adj,vis);
            if(ans!=-1) return ans*num;
        }
        return -1;
    }
};

并查集解法：

class Solution {
    // date: 2016-09-12     location: Santa Clara City Library
public:
    vector<double> calcEquation(vector<pair<string, string>> equations, vector<double>& values, vector<pair<string, string>> queries) {
        unordered_map<string, Node*> map;
        vector<double> res;
        for (int i = 0; i < equations.size(); i ++) {
            string s1 = equations[i].first, s2 = equations[i].second;
            if (map.count(s1) == 0 && map.count(s2) == 0) {
                map[s1] = new Node();
                map[s2] = new Node();
                map[s1] -> value = values[i];
                map[s2] -> value = 1;
                map[s1] -> parent = map[s2];
            } else if (map.count(s1) == 0) {
                map[s1] = new Node();
                map[s1] -> value = map[s2] -> value * values[i];
                map[s1] -> parent = map[s2];
            } else if (map.count(s2) == 0) {
                map[s2] = new Node();
                map[s2] -> value = map[s1] -> value / values[i];
                map[s2] -> parent = map[s1];
            } else {
                unionNodes(map[s1], map[s2], values[i], map);
            }
        }

        for (auto query : queries) {
            if (map.count(query.first) == 0 || map.count(query.second) == 0 || findParent(map[query.first]) != findParent(map[query.second]))
                res.push_back(-1);
            else
                res.push_back(map[query.first] -> value / map[query.second] -> value);
        }
        return res;
    }
    
private:
    struct Node {
        Node* parent;
        double value = 0.0;
        Node()  {parent = this;}
    };
    
    void unionNodes(Node* node1, Node* node2, double num, unordered_map<string, Node*>& map) {
        Node* parent1 = findParent(node1), *parent2 = findParent(node2);
        double ratio = node2 -> value * num / node1 -> value;
        for (auto it = map.begin(); it != map.end(); it ++)
            if (findParent(it -> second) == parent1)
                it -> second -> value *= ratio;
        parent1 -> parent = parent2;
    }
    
    Node* findParent(Node* node) {
        if (node -> parent == node)
            return node;
        node -> parent = findParent(node -> parent);
        return node -> parent;
    }
};