Leetcode Evaluate Reverse Polish Notation

Evaluate Reverse Polish Notation

Total Accepted:5898Total Submissions:31322My Submissions

Evaluate the value of an arithmetic expression inReverse 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

本题就是考查栈的应用。

注意细节，很容易过。 3星级的题目吧。

1 要push入结果

2 两个数的计算顺序不能错，否则答案错误

class Solution {
public:
	int evalRPN(vector<string> &tokens) 
	{
		if (tokens.size() < 1) return 0;
		stack<int> stk;

		for (int i = 0; i < tokens.size(); i++)
		{
			if (!isOperand(tokens[i])) stk.push(stoi(tokens[i]));
			else
			{
				int b = stk.top();
				stk.pop();
				int a = stk.top();
				stk.pop();
				int result = evaluate(a, b, tokens[i]);
				stk.push(result);
			}
		}
		return stk.top();
	}

	bool isOperand(const string &s) const
	{
		if (s == "+" || s == "-" || s == "*" || s == "/") return true;
		return false;
	}

	int evaluate(int a, int b, string &op)
	{
		if (op == "+") return a+b;
		if (op == "-") return a-b;
		if (op == "*") return a*b;
		return a/b;
	}
};

//2014-2-19 update
	int evalRPN(vector<string> &tokens)
	{
		stack<int> stk;
		for (int i = 0; i < tokens.size(); i++)
		{
			if (isOperand(tokens[i]))
			{
				int b = stk.top(); stk.pop();
				int a = stk.top(); stk.pop();
				int t = evaluate(tokens[i], a, b);//a b 顺序不能搞错！且需要push入结果
				stk.push(t);
			}
			else stk.push(stoi(tokens[i]));
		}
		return stk.top();
	}
	bool isOperand(string &c)
	{
		return c == "+" || c == "-" || c == "*" || c == "/";
	}
	int evaluate(string &op, int a, int b)
	{
		if (op == "+") return a+b;
		if (op == "-") return a-b;
		if (op == "*") return a*b;
		return a/b;
	}