New Problem "Expression Add Operators"

algorithms/cpp/expressionAddOperators/ExpressionAddOperators.cpp

@@ -0,0 +1,81 @@
+// Source : https://leetcode.com/problems/expression-add-operators/
+// Author : Hao Chen
+// Date   : 2016-01-16
+
+/*************************************************************************************** 
+ *
+ * Given a string that contains only digits 0-9 and a target value, return all 
+ * possibilities to add binary operators (not unary) +, -, or * between the digits so 
+ * they evaluate to the target value.
+ * 
+ * Examples: 
+ * "123", 6 -> ["1+2+3", "1*2*3"] 
+ * "232", 8 -> ["2*3+2", "2+3*2"]
+ * "105", 5 -> ["1*0+5","10-5"]
+ * "00", 0 -> ["0+0", "0-0", "0*0"]
+ * "3456237490", 9191 -> []
+ * 
+ * Credits:Special thanks to @davidtan1890 for adding this problem and creating all 
+ * test cases.
+ ***************************************************************************************/
+
+
+class Solution {
+public:
+    vector<string> addOperators(string num, int target) {
+        vector<string> result;
+        if (num.size() == 0) return result;
+        helper(num, target, result, "", 0, 0, 0, ' ');
+        return result;        
+    }
+    
+    //DFS algorithm
+    void helper(const string &num, const int target, //`num` and `target` never change
+                vector<string>& result, // the array store all of the answers
+                string solution, //the current potential answer.
+                int idx, // the current index of `num` array
+                long long val, // the current value we calculated so far
+                long long prev, // the lastest value we used for calculation, which used for operation prioirty adjustment
+                char preop ) // the latest "+" or "-" operation, which used for operation prioirty adjustment 
+    {
+        
+        if (target == val && idx == num.size()){
+            result.push_back(solution);
+            return;
+        }
+        if (idx == num.size()) return;
+        
+        string n;
+        long long v=0;
+        for(int i=idx; i<num.size(); i++) {
+            //get rid of the number which start by "0"
+            //e.g.  "05" is not the case.
+            if (n=="0") return;
+            
+            n = n + num[i];
+            v = v*10 + num[i]-'0';
+            
+            if (solution.size()==0){ 
+                // the first time for initialization
+                helper(num, target, result, n, i+1, v, 0, ' ');
+            }else{
+                // '+' or '-' needn't to adjust the priority
+                helper(num, target, result, solution + '+' + n, i+1, val+v, v, '+');
+                helper(num, target, result, solution + '-' + n, i+1, val-v, v, '-');
+                
+                //if we meet multiply operation, we need adjust the calcualtion priority
+                // e.g. if the previous value is calculated by 2+3=5, 
+                //      then if we need to multipy 4, it is not 5*4, it is 2+3*4=2+12=24
+                //      we need be careful about multiply again, such as: 2+3*4*5
+                if (preop=='+') {
+                    helper(num, target, result, solution + '*' + n, i+1, (val-prev)+prev*v, prev*v, preop);
+                }else if (preop=='-'){
+                    helper(num, target, result, solution + '*' + n, i+1, (val+prev)-prev*v, prev*v, preop);
+                }else {
+                    helper(num, target, result, solution + '*' + n, i+1, val*v, v, '*');
+                }
+            }
+        }
+        
+    }
+};