C++實現(xiàn)LeetCode(150.計算逆波蘭表達(dá)式)

更新時間：2021年07月29日 14:30:15 作者：Grandyang

這篇文章主要介紹了C++實現(xiàn)LeetCode(150.計算逆波蘭表達(dá)式),本篇文章通過簡要的案例,講解了該項技術(shù)的了解與使用,以下就是詳細(xì)內(nèi)容,需要的朋友可以參考下

[LeetCode] 150.Evaluate Reverse Polish Notation 計算逆波蘭表達(dá)式

Evaluate the value of an arithmetic expression in Reverse Polish Notation.

Valid operators are +, -, *, /. Each operand may be an integer or another expression.

Note:

Division between two integers should truncate toward zero.
The given RPN expression is always valid. That means the expression would always evaluate to a result and there won't be any divide by zero operation.

Example 1:

Input: ["2", "1", "+", "3", "*"]
Output: 9
Explanation: ((2 + 1) * 3) = 9

Example 2:

Input: ["4", "13", "5", "/", "+"]
Output: 6
Explanation: (4 + (13 / 5)) = 6

Example 3:

Input: ["10", "6", "9", "3", "+", "-11", "*", "/", "*", "17", "+", "5", "+"]
Output: 22
Explanation:
((10 * (6 / ((9 + 3) * -11))) + 17) + 5
= ((10 * (6 / (12 * -11))) + 17) + 5
= ((10 * (6 / -132)) + 17) + 5
= ((10 * 0) + 17) + 5
= (0 + 17) + 5
= 17 + 5
= 22

逆波蘭表達(dá)式就是把操作數(shù)放前面，把操作符后置的一種寫法，我們通過觀察可以發(fā)現(xiàn)，第一個出現(xiàn)的運(yùn)算符，其前面必有兩個數(shù)字，當(dāng)這個運(yùn)算符和之前兩個數(shù)字完成運(yùn)算后從原數(shù)組中刪去，把得到一個新的數(shù)字插入到原來的位置，繼續(xù)做相同運(yùn)算，直至整個數(shù)組變?yōu)橐粋€數(shù)字。于是按這種思路寫了代碼如下，但是拿到OJ上測試，發(fā)現(xiàn)會有Time Limit Exceeded的錯誤，無奈只好上網(wǎng)搜答案，發(fā)現(xiàn)大家都是用棧做的。仔細(xì)想想，這道題果然應(yīng)該是棧的完美應(yīng)用啊，從前往后遍歷數(shù)組，遇到數(shù)字則壓入棧中，遇到符號，則把棧頂?shù)膬蓚€數(shù)字拿出來運(yùn)算，把結(jié)果再壓入棧中，直到遍歷完整個數(shù)組，棧頂數(shù)字即為最終答案。代碼如下:

解法一：

class Solution {
public:
    int evalRPN(vector<string>& tokens) {
        if (tokens.size() == 1) return stoi(tokens[0]);
        stack<int> st;
        for (int i = 0; i < tokens.size(); ++i) {
            if (tokens[i] != "+" && tokens[i] != "-" && tokens[i] != "*" && tokens[i] != "/") {
                st.push(stoi(tokens[i]));
            } else {
                int num1 = st.top(); st.pop();
                int num2 = st.top(); st.pop();
                if (tokens[i] == "+") st.push(num2 + num1);
                if (tokens[i] == "-") st.push(num2 - num1);
                if (tokens[i] == "*") st.push(num2 * num1);
                if (tokens[i] == "/") st.push(num2 / num1);
            }
        }
        return st.top();
    }
};

我們也可以用遞歸來做，由于一個有效的逆波蘭表達(dá)式的末尾必定是操作符，所以我們可以從末尾開始處理，如果遇到操作符，向前兩個位置調(diào)用遞歸函數(shù)，找出前面兩個數(shù)字，然后進(jìn)行操作將結(jié)果返回，如果遇到的是數(shù)字直接返回即可，參見代碼如下：

解法二：

class Solution {
public:
    int evalRPN(vector<string>& tokens) {
        int op = (int)tokens.size() - 1;
        return helper(tokens, op);
    }
    int helper(vector<string>& tokens, int& op) {
        string str = tokens[op];
        if (str != "+" && str != "-" && str != "*" && str != "/") return stoi(str);
        int num1 = helper(tokens, --op);
        int num2 = helper(tokens, --op);
        if (str == "+") return num2 + num1;
        if (str == "-") return num2 - num1;
        if (str == "*") return num2 * num1;
        return num2 / num1;
    }
};

類似題目：

Basic Calculator

Expression Add Operators

參考資料：

https://leetcode.com/problemset/algorithms/

https://leetcode.com/problems/evaluate-reverse-polish-notation/discuss/47642/a-recursive-solution-in-cpp

https://leetcode.com/problems/evaluate-reverse-polish-notation/discuss/47544/Challenge-me-neat-C%2B%2B-solution-could-be-simpler

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