实际为图的遍历问题,图采用邻接表的形式存储(用unordered_map存),键为图中的节点,值为与该节点直接相连的其余节点(包含边的权重,边的权重就是除的结果)。好比a/b=2,就将a存为键,b和2存为值,同时b也为节点,a和1/2存为值。假设已知a/b=2,b/c=3,求a/c,那么就是求a点到c点的路径(将路径上的权值相乘,即2*3=6)。故本质为图的遍历。web
以前博客关于图的遍历的文章:https://blog.csdn.net/aikudexue/article/details/89684046svg
1.DFS
参考:
https://blog.csdn.net/KID_LWC/article/details/81276050lua
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; //用邻接表方式保存图 unordered_set<string> vis; //存储已经访问的节点 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++){ //对要查结果中的每一对点对 string s1=queries[i].first,s2=queries[i].second; //s1为起点,s2为终点 double val=dfs(s1,s2,adj,vis); //从图adj中找s1->s2的路径,已经访问过的点保存在vis里,返回结果为s1->s2的边权值乘积 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; //若是找的是本身自己,则本身到本身的权值为1 if(vis.find(s1)!=vis.end()) return -1; //若是s1已经访问过了,即已经在vis当中了则直接返回没找到 vis.insert(s1); //把s1放入vis里 for(auto itr=adj[s1].begin();itr!=adj[s1].end();++itr){ //下次找和s1直接相连的全部元素 string next = itr->first; //下一步从next节点开始找s2,至关于s1的子树 double num = itr->second; double ans = dfs(next,s2,adj,vis); //从next开始找s2 if(ans!=-1) return ans*num; //s1->s2的权值 = next->s2的权值*s1->next的权值 } return -1; } };
leetcode中DFS的代码:
https://leetcode.com/problems/evaluate-division/discuss/88168/c%2B%2B-0ms-Hash%2BDFS-solutionspa
2.BFS
参考:
https://leetcode.com/problems/evaluate-division/discuss/297938/C%2B%2B-BFS-%2B-Memoization(for-O(1)-query-retrieval).net
https://www.jianshu.com/p/b7ab2384bed1
(还没看,随便找的一个答案)code
3.并查集(没看)
参考:
https://leetcode.com/problems/evaluate-division/discuss/147281/Java-Union-Find-solution-faster-than-99xml