[leetcode]399. Evaluate Division

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本文介绍了一种解决LeetCode上分式求值问题的方法。该问题涉及通过一系列给定的等式（如 a/b=2.0, b/c=3.0）来解答特定变量之间的比例关系。文章详细解释了如何利用哈希表存储每个变量对应的值，并通过统一基空间的方式处理不同组的变量关系，从而高效地解决查询问题。

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题目链接：https://leetcode.com/problems/evaluate-division/

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.

解题思路：初始化mp,base,mp存储每一个字母到数字的映射，base存储每一个字母所属的“基空间”，比如a/b=1.0，b/c=2.0，e/f=3.0，可以看到，a、b、c是有联系的，因此可以假定a、b、c的"基空间"为a，当然也可以假定为b或c；e、f这两个是联系的，但与a、b、c是不联系的，所以可以假定e、f的“基空间”为e或f。

遍历equations，可以出现四种情况(设被除数是s1，除数是s2,商是v):

情况一：s1和s2在mp中都没有出现过，让mp[s1]=v, mp[s2]=1, base[s1]=s1, base[s2]=s1;

情况二：s1在mp中没有出现过，让mp[s1]=mp[s2]*v, base[s1]=s2;

情况三：s2在mp中没有出现过，让mp[s2]=mp[s1]/v, base[s2]=s1;

情况四：s1和s2均在mp中出现过，调整s1和s2的“基空间”，使s1和s2属于同一个“基空间”，假设把所有的“基空间”调整成s1所属的“基空间”,那么调整倍率为mp[s2]*v/mp[s1]。接着遍历整个mp，使属于s1的“基空间”的字母乘上这个倍率。

最后，根据要查询的字母返回相应的值。

class Solution{
public:
    vector<double> calcEquation(vector<pair<string,string>> equations,vector<double>& values,vector<pair<string,string>> queries)
    {
        vector<double> res;
        unordered_map<string,double> mp;
        for(int i=0;i<equations.size();i++)
        {
            if(mp.count(equations[i].first)==0 && mp.count(equations[i].second)==0)
            {
                mp[equations[i].first]=values[i];
                mp[equations[i].second]=1;
                base[equations[i].first]=equations[i].first;
                base[equations[i].second]=equations[i].first;
            }
            else if(mp.count(equations[i].first)==0)
            {
                mp[equations[i].first]=mp[equations[i].second]*values[i];
                base[equations[i].first]=equations[i].second;
            }
            else if(mp.count(equations[i].second)==0)
            {
                mp[equations[i].second]=mp[equations[i].first]/values[i];
                base[equations[i].second]=equations[i].first;
            }
            else
                unionBase(equations[i].first,equations[i].second,values[i],mp);
        }
        for(int i=0;i<queries.size();i++)
        {
            if(mp.count(queries[i].first)==0 || mp.count(queries[i].second)==0 || findBase(queries[i].first)!=findBase(queries[i].second))
                res.push_back(-1);
            else
                res.push_back(mp[queries[i].first]/mp[queries[i].second]);
        }
        return res;
    }

    void unionBase(string s1,string s2,double v,unordered_map<string,double>& mp)
    {
        string base1=findBase(s1),base2=findBase(s2);
        double adjustF=mp[s2]*v/mp[s1];
        for(auto it=mp.begin();it!=mp.end();++it)
        {
            if(findBase(it->first)==base1)
                mp[it->first]*=adjustF;
        }
        base[s2]=s1;
    }

    string findBase(string s)
    {
        while(base[s]!=s) s=base[s];
        return s;
    }
private:
    unordered_map<string,string> base;

};