Evaluate Reverse Polish Notation
Evaluate the value of an arithmetic expression in Reverse Polish Notation.
Valid operators are
+
, -
, *
, /
. Each operand may be an integer or another expression.
Some examples:
["2", "1", "+", "3", "*"] -> ((2 + 1) * 3) -> 9 ["4", "13", "5", "/", "+"] -> (4 + (13 / 5)) -> 6
------------------------ thinking ---------------------------------------------------
using stack. The main thinking is right. But there are a lot of bugs during the implementation!!!
1. stack.back() and stack.pop_back() should be call in sequential!!
2. the tokens[i]'s size should be considered!! we have chances to get "+2" or "-4"
3. after if for size() > 1 checking, we should break!
--------------------------- codes ----------------------------------------------------
class Solution { public: int evalRPN(vector<string> &tokens) { vector<int> stack; for (int i = 0; i < tokens.size(); i++) { if (tokens[i].size() > 1) { //bug here stack.push_back(stoi(tokens[i])); } else {//bug here switch (tokens[i][0]) { case '+': { int a = stack.back(); stack.pop_back(); int b = stack.back(); stack.pop_back(); stack.push_back(a + b); break; } case '-': { int a = stack.back(); stack.pop_back(); int b = stack.back(); stack.pop_back(); stack.push_back(b - a); break; } case '*': { int a = stack.back(); stack.pop_back(); int b = stack.back(); stack.pop_back(); stack.push_back(a * b); break; } case '/': { int a = stack.back(); stack.pop_back(); int b = stack.back(); stack.pop_back(); stack.push_back(b / a); break; } default: { int a = stoi(tokens[i]); stack.push_back(a); break; } } } } return stack.back(); } };
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