leetcode练习 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

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.

代码：

class Solution {
public:
    vector<double> calcEquation(vector<pair<string, string>> equations, vector<double>& values, vector<pair<string, string>> queries) {
        vector<double> result;
        vector<bool> visit;
        vector<string> nodes;
        for (int i = 0; i < equations.size(); i++) {
            visit.push_back(false);
            nodes.push_back(equations[i].first);
            nodes.push_back(equations[i].second);
        }
        for (int i = 0; i < queries.size(); i++){
            string src = queries[i].first;
            string dst = queries[i].second;
            double r = 1;
            r = r * findPath(src, dst, equations, visit, values);
            if (r == 0) r = -1;
            int flag = 0;
            for (int j = 0; j < nodes.size(); j++) {
                if (nodes[j] == src) {
                    flag = 1;
                    break;
                }
            }
            if (src == dst && flag) r = 1;
            result.push_back(r);
        }
        return result;

    }
    double findPath(string s, string d, vector<pair<string, string>> equations, vector<bool> visit, vector<double> values){
        double r = 0;
        double rm = 0;
        string src = s;
        string dst = d;
        for (int i = 0; i < equations.size(); i++) {
            if (equations[i].first == s && equations[i].second == d && visit[i] == false) {
                r = 1;
                visit[i] = true;
                return values[i];
            }
            else if (equations[i].first == d && equations[i].second == s && visit[i] == false) {
                r = 1;
                visit[i] = true;
                return 1 / values[i];
            }
            else if (equations[i].first == s && visit[i] == false) {
                r = 1;
                visit[i] = true;
                r = values[i];
                src = equations[i].second;
                r = r*findPath(src, dst, equations, visit, values);
            }
            else if (equations[i].second == s && visit[i] == false) {
                r = 1;
                visit[i] = true;
                r = 1 / values[i];
                src = equations[i].first;
                r = r*findPath(src, dst, equations, visit, values);
            }
            if (r > 0) rm = r;
        }
        return rm;
    }
};

用了深度优先搜索，感受到了自己对于这块知识的陌生，小的细节思考了非常久。比如对于没找到目标的分支的处理非常不得心应手，还需要锻炼。