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 程式師世界 >> 編程語言 >> C語言 >> C++ >> 關於C++ >> [LeetCode] Basic Calculator II

[LeetCode] Basic Calculator II

編輯:關於C++

Basic Calculator II

Implement a basic calculator to evaluate a simple expression string.

The expression string contains only non-negative integers, +, -, *, and / operators. 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.

解題思路:

同Basic Calculator I一樣,只需改動轉化成後綴表達式的代碼。

class Solution {
public:
    int calculate(string s) {
        string postS = getPostString(s);    //獲得後綴表達式
        cout << postS;
        stack nums;
        int len = postS.length();
        int num1, num2;
        for (int i = 0; i= '0' && postS[i] <= '9'){
                    num = num * 10 + (postS[i] - '0');
                    i++;
                }
                i--;
                nums.push(num);
                break;
            }
        }
        return nums.top();
    }
  
private:
    string getPostString(string s){
        string postS = "";
  
        stack op;
        int len = s.length();
        for (int i = 0; i= '0'&&s[i] <= '9'){
                    postS += s[i];
                    i++;
                }
                postS += '#';   //分隔數字
                i--;
                break;
            }
        }
        while (!op.empty()){
            postS += op.top();
            op.pop();
        }
  
        return postS;
    }
};

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