LintCode 424: Evaluate Reverse Polish Notation 递归和栈好题

424 · Evaluate Reverse Polish Notation
Description
Evaluate the value of an arithmetic expression in Reverse Polish Notation.
Valid operators are +, -, , /. Each operand may be an integer or another expression.
If you use python, you should use int(a / b) to calc the / operation rather than a // b.
Example
Example 1:
Input: [“2”, “1”, “+”, “3”, ""]
Output: 9
Explanation: [“2”, “1”, “+”, “3”, “*”] -> (2 + 1) * 3 -> 9
Example 2:
Input: [“4”, “13”, “5”, “/”, “+”]
Output: 6
Explanation: [“4”, “13”, “5”, “/”, “+”] -> 4 + 13 / 5 -> 6

解法1：递归
以[“5”, “3”, “4”, “-”, “+”]为例，一开始第0层调用helper(tokens, 4)，从右往左扫描，遇到”+“号，先第一层递归调用helper(tokens, 3)，这里会遇到"-"，第二层递归调用helper(tokens,2)，此时遇到4和3,得到3-4=-1，然后再递归调用helper(tokens, 1)，得到5，然后得到5+(-1)=4。注意递归参数op要用引用，因为每次递归都会-1。

class Solution {
public:
    /**
     * @param tokens: The Reverse Polish Notation
     * @return: the value
     */
    int evalRPN(vector<string> &tokens) {
        int op = tokens.size() - 1;
        return helper(tokens, op);
    }
private:
    int helper(vector<string>& tokens, int& op) {
        string s = tokens[op];
        if (s != "+" && s != "-" && s != "*" && s != "/") return stoi(s);
        int v1 = helper(tokens, --op);
        int v2 = helper(tokens, --op);
        if (s == "+") return v1 + v2;
        if (s == "-") return v2 - v1;
        if (s == "*") return v1 * v2;
        return v2 / v1;
    }
};

解法2：用stack，这个解法最好理解。
以[“5”, “3”, “4”, “-”, “+”]为例，，从左往右扫描，5，3，4先顺序压栈，遇到"-“，弹出4和3，然后把3-4的结果-1压栈，然后遇到”+"，弹出-1和5，得到5+(-1)的结果为4。

class Solution {
public:
    /**
     * @param tokens: The Reverse Polish Notation
     * @return: the value
     */
    int evalRPN(vector<string> &tokens) {
        int len = tokens.size();
        stack<int> stk;
        for (int i = 0; i < len; i++) {
            string s = tokens[i];
            if (s != "+" && s != "-" && s != "*" && s != "/") {
                stk.push(stoi(s));
                continue;
            }
            int result = 0;
            int top1 = stk.top(); stk.pop();
            int top2 = stk.top(); stk.pop();
            switch (s[0]) {
                case '+': 
                    result = top1 + top2;
                    break;
                case '-': 
                    result = top2 - top1;
                    break;
                case '*': 
                    result = top1 * top2;
                    break;
                case '/': 
                    result = top2 / top1;
                    break;
                default:
                    break;
            }
            stk.push(result);
        }        
        int ret = stk.top(); stk.pop();
        return ret;                    
    }
};