[LeetCode] 227. Basic Calculator II 基本計算器之二

 

Implement a basic calculator to evaluate a simple expression string.

The expression string contains only non-negativeintegers, +, -, *, / operators and empty spaces . The integer division should truncate toward zero.

Example 1:

Input: "3+2*2"
Output: 7

Example 2:

Input: " 3/2 "
Output: 1

Example 3:

Input: " 3+5 / 2 "
Output: 5

Note:

• You may assume that the given expression is always valid.
• Do not use the eval built-in library function.

Credits:
Special thanks to @ts for adding this problem and creating all test cases.

 

這道題是之前那道 Basic Calculator 的拓展，不同之處在於那道題的計算符號只有加和減，而這題加上了乘除，那么就牽扯到了運算優先級的問題，好在這道題去掉了括號，還適當的降低了難度，估計再出一道的話就該加上括號了。不管那么多，這道題先按木有有括號來處理，由於存在運算優先級，我們采取的措施是使用一個棧保存數字，如果該數字之前的符號是加或減，那么把當前數字壓入棧中，注意如果是減號，則加入當前數字的相反數，因為減法相當於加上一個相反數。如果之前的符號是乘或除，那么從棧頂取出一個數字和當前數字進行乘或除的運算，再把結果壓入棧中，那么完成一遍遍歷后，所有的乘或除都運算完了，再把棧中所有的數字都加起來就是最終結果了，參見代碼如下：

 

解法一：

class Solution {
public:
    int calculate(string s) {
        long res = 0, num = 0, n = s.size();
        char op = '+';
        stack<int> st;
        for (int i = 0; i < n; ++i) {
            if (s[i] >= '0') {
                num = num * 10 + s[i] - '0';
            }
            if ((s[i] < '0' && s[i] != ' ') || i == n - 1) {
                if (op == '+') st.push(num);
                if (op == '-') st.push(-num);
                if (op == '*' || op == '/') {
                    int tmp = (op == '*') ? st.top() * num : st.top() / num;
                    st.pop();
                    st.push(tmp);
                }
                op = s[i];
                num = 0;
            } 
        }
        while (!st.empty()) {
            res += st.top();
            st.pop();
        }
        return res;
    }
};

 

在做了 Basic Calculator III 之后，再反過頭來看這道題，發現只要將處理括號的部分去掉直接就可以在這道題上使用，參見代碼如下：

 

解法二：

class Solution {
public:
    int calculate(string s) {
        long res = 0, curRes = 0, num = 0, n = s.size();
        char op = '+';
        for (int i = 0; i < n; ++i) {
            char c = s[i];
            if (c >= '0' && c <= '9') {
                num = num * 10 + c - '0';
            }
            if (c == '+' || c == '-' || c == '*' || c == '/' || i == n - 1) {
                switch (op) {
                    case '+': curRes += num; break;
                    case '-': curRes -= num; break;
                    case '*': curRes *= num; break;
                    case '/': curRes /= num; break;
                }
                if (c == '+' || c == '-' || i == n - 1) {
                    res += curRes;
                    curRes = 0;
                }
                op = c;
                num = 0;
            } 
        }
        return res;
    }
};

 

Github 同步地址：

https://github.com/grandyang/leetcode/issues/227

 

類似題目：

Basic Calculator III

Basic Calculator

Expression Add Operators

 

參考資料：

https://leetcode.com/problems/basic-calculator-ii/

https://leetcode.com/problems/basic-calculator-ii/discuss/63003/Share-my-java-solution

https://leetcode.com/problems/basic-calculator-ii/discuss/63004/17-lines-C++-easy-20-ms

https://leetcode.com/problems/basic-calculator-ii/discuss/63031/Simple-C++-solution-beats-85-submissions-with-detailed-explanations

 

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