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Author: assassint2017

Evaluate Reverse Polish Notation/Evaluate_Reverse_Polish_Notation.cpp

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+// 典型的利用堆栈的题目
+// Runtime: 16 ms, faster than 100.00% of C++ online submissions for Evaluate Reverse Polish Notation.
+// Memory Usage: 11.5 MB, less than 39.61% of C++ online submissions for Evaluate Reverse Polish Notation.
+class Solution 
+{
+public:
+    int evalRPN(vector<string>& tokens) 
+    {
+        stack<int> memo;
+        int res;
+
+        for (auto token : tokens)
+        {
+            char chr = token[0];
+
+            // 如果是数字的话就压入堆栈
+            if (chr >= '0' && chr <= '9' || token.size() > 1)
+                memo.push(atoi(token.c_str()));
+                
+            // 如果是运算符的话就要从堆栈中提取出两个运算数进行运算
+            else
+            {
+                int op1 = memo.top();
+                memo.pop();
+                int op2 = memo.top();
+                memo.pop();
+
+                if (chr == '+')
+                    res = op2 + op1;
+                else if (chr == '-')
+                    res = op2 - op1;
+                else if (chr == '*')
+                    res = op2 * op1;
+                else if (chr == '/')
+                    res = op2 / op1;
+                
+                memo.push(res);
+            }
+        }
+        return memo.top();
+    }
+};

Evaluate Reverse Polish Notation/Evaluate_Reverse_Polish_Notation.py

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+# 这里要注意python中的 // 运算符的问题所在
+
+# Runtime: 36 ms, faster than 100.00% of Python3 online submissions for Evaluate Reverse Polish Notation.
+# Memory Usage: 12.9 MB, less than 24.06% of Python3 online submissions for Evaluate Reverse Polish Notation.
+
+class Solution:
+    def evalRPN(self, tokens: 'List[str]') -> 'int':
+        memo = []
+
+        for token in tokens:
+            if '0' <= token <= '9' or len(token) > 1:
+                memo.append(int(token))
+            else:
+                op1 = memo.pop()
+                op2 = memo.pop()
+
+                if token == '+':
+                    memo.append(op2 + op1)
+                elif token == '-':
+                    memo.append(op2 - op1)
+                elif token == '*':
+                    memo.append(op2 * op1)
+                else:
+                    memo.append(int(op2 / op1))
+                
+        return memo.pop();