LintCode 1908: Boolean expression evaluation (栈好题)

1908 · Boolean expression evaluation
Algorithms
Hard
Accepted Rate
37%
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Description
Given a string represents a Boolean expression only including “true”,“false”,“or”,“and”.
Your task is to calculate the result of the expression by returning “true” or “false”.
If the expression is not valid, you need to return a string “error”.

We promise that there are no more than 1000010000 elements in the expression.
There are only 4 kinds of elements in the expression : “true”, “false”, “and”, “or”.

Example
Example 1

Input：
“true and false”
Output：
“false”
Example 2

Input：
“true or”
Output：
“error”
Tags
Company
Meituan

解法1：用stack, 类似LintCode 368 Expression Evaluate那题，中序表达式转RPN后求值。注意LintCode 368那题不需要检查表达式是否有错，这题需要，所以就判断"and"和"or“前后是不是"true"和"false"即可。

class Solution {
public:
    /**
     * @param expression: a string that representing an expression
     * @return: the result of the expression
     */
    string evaluation(string &expression) {
        stack<string> optrStk;
        vector<string> RPNExpr;
        map<string, int> prio;
        prio["and"] = 2;
        prio["or"] = 1;
        prio["true"] = 0;
        prio["false"] = 0;
        //tokenize
        stringstream ss(expression);
        vector<string> tokens;
        string token;
        while (ss >> token) tokens.push_back(token);
        int len = tokens.size();
        
        //transform it to RPN
        for (int i = 0; i < len; i++) 
        {
            string expr = tokens[i];
            if (expr == "true" || expr == "false") {
                RPNExpr.push_back(expr);
                continue;
            }
            
            if (i == 0 || i == len - 1) return "error";  //"and" or "or" is the first token or last token, error!
            if (prio[tokens[i - 1]] > 0 || prio[tokens[i + 1]] > 0) return "error";
            while (!optrStk.empty() && prio[optrStk.top()] >= prio[expr]) {
                RPNExpr.push_back(optrStk.top());
                optrStk.pop();
            }
            optrStk.push(expr);
        }
        //dump the rest in optrStr to RPNExpr
        while (!optrStk.empty()) {
            RPNExpr.push_back(optrStk.top());
            optrStk.pop();
        }

        //now evaluate RPN
        bool ret;
        len = RPNExpr.size();
        stack<bool> RPNStk;
        if (len == 0 || len == 2) return "error";
        for (int i = 0; i < len; i++) {
            string expr = RPNExpr[i];
            if (expr == "true" || expr == "false") {
                RPNStk.push(expr == "true");
            } else {
                bool first = RPNStk.top(); RPNStk.pop();
                bool second = RPNStk.top(); RPNStk.pop();
               
                if (expr == "and") {
                    RPNStk.push(first && second);
                } else  { //expr == "or"
                    RPNStk.push(first || second);
                }
            }
        }
        if (RPNStk.top()) return "true";
        return "false";
    }
};

解法2：这题应该也可以用递归解，先找出优先级最低的那个符号，然后左右递归求值。