add prob #150; O(N) in time and O(N) in space;

Author: randomwalk10

150_Evaluate_Reverse_Polish_Notation.cpp

@@ -0,0 +1,86 @@
+/*
+Evaluate the value of an arithmetic expression in Reverse Polish Notation.
+
+Valid operators are +, -, *, /. Each operand may be an integer or another expression.
+
+Note:
+
+Division between two integers should truncate toward zero.
+The given RPN expression is always valid. That means the expression would always evaluate to a result and there won't be any divide by zero operation.
+Example 1:
+
+Input: ["2", "1", "+", "3", "*"]
+Output: 9
+Explanation: ((2 + 1) * 3) = 9
+Example 2:
+
+Input: ["4", "13", "5", "/", "+"]
+Output: 6
+Explanation: (4 + (13 / 5)) = 6
+Example 3:
+
+Input: ["10", "6", "9", "3", "+", "-11", "*", "/", "*", "17", "+", "5", "+"]
+Output: 22
+Explanation: 
+  ((10 * (6 / ((9 + 3) * -11))) + 17) + 5
+= ((10 * (6 / (12 * -11))) + 17) + 5
+= ((10 * (6 / -132)) + 17) + 5
+= ((10 * 0) + 17) + 5
+= (0 + 17) + 5
+= 17 + 5
+= 22
+
+来源：力扣（LeetCode）
+链接：https://leetcode-cn.com/problems/evaluate-reverse-polish-notation
+著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。*/
+
+/* O(N) in time and O(N) in space */
+#include <vector>
+#include <string>
+#include <stack>
+using namespace std;
+
+class Solution {
+public:
+    int evalRPN(vector<string>& tokens) {
+        stack<int> s;
+
+        for(int i=0; i<(int)tokens.size(); ++i){
+            if(tokens[i][0]=='+'){
+                int rhs = s.top(); s.pop();
+                int lhs = s.top(); s.pop();
+                s.push(lhs+rhs);
+            }
+            else if( (tokens[i].size()==1)&&(tokens[i][0]=='-') ){
+                int rhs = s.top(); s.pop();
+                int lhs = s.top(); s.pop();
+                s.push(lhs-rhs);
+            }
+            else if(tokens[i][0]=='*'){
+                int rhs = s.top(); s.pop();
+                int lhs = s.top(); s.pop();
+                s.push(lhs*rhs);
+            }
+            else if(tokens[i][0]=='/'){
+                int rhs = s.top(); s.pop();
+                int lhs = s.top(); s.pop();
+                s.push(lhs/rhs);
+            }
+            else{
+                int temp = 0;
+                bool is_pos = true;
+                for(int j=0; j<(int)tokens[i].size(); ++j){
+                    if(tokens[i][j]=='-') is_pos = false;
+                    else{
+                        temp = temp*10 + tokens[i][j] - '0';
+                    }
+                }
+
+                if(is_pos) s.push(temp);
+                else s.push(-temp);
+            }
+        }
+
+        return s.top();
+    }
+};