[LeetCode] Basic Calculator II 基本计算器之二

 

Implement a basic calculator to evaluate a simple expression string.

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

You may assume that the given expression is always valid.

Some examples:

"3+2*2" = 7
" 3/2 " = 1
" 3+5 / 2 " = 5

Note: Do not use the eval built-in library function.

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

 

这道题是之前那道Basic Calculator 基本计算器的拓展，不同之处在于那道题的计算符号只有加和减，而这题加上了乘除，那么就牵扯到了运算优先级的问题，好在这道题去掉了括号，还适当的降低了难度，估计再出一道的话就该加上括号了。不管那么多，这道题先按木有有括号来处理，由于存在运算优先级，我们采取的措施是使用一个栈保存数字，如果该数字之前的符号是加或减，那么把当前数字压入栈中，注意如果是减号，则加入当前数字的相反数，因为减法相当于加上一个相反数。如果之前的符号是乘或除，那么从栈顶取出一个数字和当前数字进行乘或除的运算，再把结果压入栈中，那么完成一遍遍历后，所有的乘或除都运算完了，再把栈中所有的数字都加起来就是最终结果了。代码如下：

 

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

 

参考资料：

https://leetcode.com/discuss/42423/my-28ms-c-code-with-two-stacks-one-for-op-one-for-oprand

https://leetcode.com/discuss/41641/17-lines-c-easy-20-ms

https://leetcode.com/discuss/41902/share-my-java-solution

 

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