題目描述

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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

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最初代碼：

1. 最初沒好好看題，沒注意里面的數(shù)字可能是整數(shù)也可能是表達式，思考簡單了。一直說越界或者其他。一直想為什么用string，不是char，原來會有表達式。真的很菜..

2. 理解之后就可以寫了

int evalRPN(vector<string> &tokens) {stack<int> numbers;int n = tokens.size();if(n == 0) return 0;for(int i = 0; i < tokens.size(); i++){if(tokens[i] >= "0" && tokens[i] <= "9"){numbers.push(stoi(tokens[i]));}else{int b = numbers.top();numbers.pop();int a = numbers.top();numbers.pop();int result = 0;if(tokens[i] == "+")result = a + b;else if(tokens[i] == "-")result = a - b;else if(tokens[i] == "*")result = a * b;else if(tokens[i] == "/")result = a / b;numbers.push(result);}}return numbers.top(); }

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AC代碼：

class Solution { public:int evalRPN(vector<string> &tokens) {stack<int> numbers;int n = tokens.size();if(n == 0) return 0;for(auto x : tokens){if(x == "+" || x == "-" || x == "*" || x == "/"){if(numbers.size() < 2) return 0;int b = numbers.top();numbers.pop();int a = numbers.top();numbers.pop();if(x == "+") numbers.push(a+b);if(x == "-") numbers.push(a-b); if(x == "*") numbers.push(a*b);if(x == "/") numbers.push(a/b);}else{numbers.push(atoi(x.c_str()));}}return numbers.top();} };

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