leetcode-399: division evaluation

题目

题目连接

给你一个变量对数组 equations 和一个实数值数组 values 作为已知条件,其中 equations[i] = [Ai, Bi] 和 values[i] 共同表示等式 Ai / Bi = values[i] .每个 Ai 或 Bi 是一个表示单个变量的字符串.

另有一些以数组 queries 表示的问题,其中 queries[j] = [Cj, Dj] 表示第 j 个问题,请你根据已知条件找出 Cj / Dj = ? 的结果作为答案.

返回 所有问题的答案 .如果存在某个无法确定的答案,则用 -1.0 替代这个答案.如果问题中出现了给定的已知条件中没有出现的字符串,也需要用 -1.0 替代这个答案.

注意：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.

示例 1：

输入：equations = [["a","b"],["b","c"]], values = [2.0,3.0], queries = [["a","c"],["b","a"],["a","e"],["a","a"],["x","x"]]
输出：[6.00000,0.50000,-1.00000,1.00000,-1.00000]
解释：
条件：a / b = 2.0, b / c = 3.0
问题：a / c = ?, b / a = ?, a / e = ?, a / a = ?, x / x = ?
结果：[6.0, 0.5, -1.0, 1.0, -1.0 ]

示例 2：

输入：equations = [["a","b"],["b","c"],["bc","cd"]], values = [1.5,2.5,5.0], queries = [["a","c"],["c","b"],["bc","cd"],["cd","bc"]]
输出：[3.75000,0.40000,5.00000,0.20000]

示例 3：

输入：equations = [["a","b"]], values = [0.5], queries = [["a","b"],["b","a"],["a","c"],["x","y"]]
输出：[0.50000,2.00000,-1.00000,-1.00000]

解题

方法一：并查集

参考链接

Building a weighted directed graph+并查集

class UnionFind{
    
private:
    vector<int> parent;
    vector<double> weight;
public:
    UnionFind(int n){
    
        parent.resize(n);
        weight.resize(n);
        for(int i=0;i<n;i++){
    
            parent[i]=i;
            weight[i]=1;
        }
    }
    int find(int index){
    
        if(index==parent[index]) return index;
        int origin=parent[index];
        parent[index]=find(parent[index]);//At the same time, the parent node will be displayedweight也更新好
        weight[index]*=weight[origin];
        return parent[index];
    }
    void unite(int  index1,int index2,double value){
    
        int p1=find(index1);
        int p2=find(index2);
        if(p1!=p2){
    
            parent[p1]=p2;
            weight[p1]=weight[index2]*value/weight[index1];
        }
    }
    double isConnected(int index1,int index2){
    
        int p1=find(index1);
        int p2=find(index2);
        if(p1==p2){
    
            return weight[index1]/weight[index2];
        }
        else return -1;
    }
};


class Solution {
    
public:
    vector<double> calcEquation(vector<vector<string>>& equations, vector<double>& values, vector<vector<string>>& queries) {
    
        int equationSize=equations.size();
        UnionFind uf(2*equationSize);
        // 第 1 步：预处理,将变量的值与 id 进行映射,使得并查集的底层使用数组实现,方便编码
        unordered_map<string,int> mp;
        int id=0;
        for(int i=0;i<equationSize;i++){
    
            vector<string> equation=equations[i];
            string s1=equation[0];
            string s2=equation[1];
            if(!mp.count(s1)){
    
                mp[s1]=id++;
            }
            if(!mp.count(s2)){
    
                mp[s2]=id++;
            }
            uf.unite(mp[s1],mp[s2],values[i]);
        }
        // 第 2 步：做查询
        int queriesSize=queries.size();
        vector<double> res(queriesSize);
        for(int i=0;i<queriesSize;i++){
    
            string s1=queries[i][0];
            string s2=queries[i][1];
            if(!mp.count(s1)||!mp.count(s2)) res[i]=-1;
            else res[i]=uf.isConnected(mp[s1],mp[s2]);
        }
        return res;
    }
};