New Problem Solution "Evaluate Division"

Author: haoel

README.md

@@ -8,6 +8,7 @@ LeetCode
 
 | # | Title | Solution | Difficulty |
 |---| ----- | -------- | ---------- |
+|399|[Evaluate Division](https://leetcode.com/problems/evaluate-division/) | [C++](./algorithms/cpp/evaluateDivision/EvaluateDivision.cpp)|Medium|
 |398|[Random Pick Index](https://leetcode.com/problems/random-pick-index/) | [C++](./algorithms/cpp/randomPickIndex/RandomPickIndex.cpp)|Medium|
 |397|[Integer Replacement](https://leetcode.com/problems/integer-replacement/) | [C++](./algorithms/cpp/integerReplacement/IntegerReplacement.cpp)|Medium|
 |396|[Rotate Function](https://leetcode.com/problems/rotate-function/) | [C++](./algorithms/cpp/rotateFunction/RotateFunction.cpp)|Easy|

algorithms/cpp/evaluateDivision/EvaluateDivision.cpp

@@ -0,0 +1,93 @@
+// Source : https://leetcode.com/problems/evaluate-division/
+// Author : Hao Chen
+// Date   : 2016-11-05
+
+/*************************************************************************************** 
+ *
+ * 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.
+ ***************************************************************************************/
+
+class Solution {
+private:
+    bool dfs( unordered_map<string, unordered_map<string, double>>& m, 
+              unordered_map<string, bool>& visited, 
+              string& start, string& end, double& res  ) {
+                 
+        if ( m.find(start) == m.end() || m.find(end) == m.end() ) return false;
+        if ( start == end ) return true;
+        
+        for (auto it = m[start].begin(); it != m[start].end(); ++it) {
+           
+            auto key = it->first;
+            auto value = it->second;
+            
+            // already visited, skip it.
+            if (visited.find(key) != visited.end() ) {
+                continue;
+            }
+            
+            visited[key] = true;
+            double old = res;
+            res *= value;
+            
+            if (dfs(m, visited, key, end, res)) {
+                return true;
+            }
+            //didn't find the result, reset the current result, and go to next one
+            res = old;
+            visited.erase(key);
+        }
+        
+        return false;
+    }
+public:
+    vector<double> calcEquation(vector<pair<string, string>> equations, 
+                                vector<double>& values, 
+                                vector<pair<string, string>> queries) {
+        
+        unordered_map<string, unordered_map<string, double>> m;
+        for(int i=0; i<equations.size(); i++) {
+           auto first = equations[i].first;
+           auto second = equations[i].second;
+           m[first][second] = values[i];
+           m[second][first] = 1.0 / values[i];
+        }
+        
+        
+        vector<double> result;
+        for(auto q : queries) {
+            string start = q.first;
+            string end = q.second;
+            
+            unordered_map<string, bool> visited;
+            visited[start] = true;
+            double res = 1.0;
+            
+            if(dfs(m, visited, start, end, res)) {
+                result.push_back(res);
+            } else {
+                result.push_back(-1.0);
+            }
+        }
+        
+        return result;
+    }
+};