題目描述

You’re now a baseball game point recorder.

Given a list of strings, each string can be one of the 4 following types:

1. Integer (one round’s score): Directly represents the number of points you get in this round.
2. "+" (one round’s score): Represents that the points you get in this round are the sum of the last two valid round’s points.
3. "D" (one round’s score): Represents that the points you get in this round are the doubled data of the last valid round’s points.
4. "C" (an operation, which isn’t a round’s score): Represents the last valid round’s points you get were invalid and should be removed.

Each round’s operation is permanent and could have an impact on the round before and the round after.

You need to return the sum of the points you could get in all the rounds.

Example 1:

Input: ["5","2","C","D","+"]
Output: 30
Explanation: 
Round 1: You could get 5 points. The sum is: 5.
Round 2: You could get 2 points. The sum is: 7.
Operation 1: The round 2's data was invalid. The sum is: 5.  
Round 3
 
: You could get 10 points (the round 2's data has been removed). The sum is: 15.
Round 4: You could get 5 + 10 = 15 points. The sum is: 30.

Example 2:

Input: ["5","-2","4","C","D","9","+","+"]
Output: 27
Explanation: 
Round 1: You could get 5 points. The sum is: 5.
Round 2: You could get -2 points. The sum is: 3.
Round 3: You could get 4 points. The sum is: 7.
Operation 1: The round 3's data is invalid. The sum is: 3.  
Round 4: You could get -4 points (the round 3's data has been removed). The sum is: -1.
Round 5: You could get 9 points. The sum is: 8.
Round 6: You could get -4 + 9 = 5 points. The sum is 13.
Round 7: You could get 9 + 5 = 14 points. The sum is 27.

Note:

• The size of the input list will be between 1 and 1000.
• Every integer represented in the list will be between -30000 and 30000.

理解題意

輸入一個字串列表，根據字串的不同會有不同的要求。

演算法

直接使用動態陣列容器vector，迴圈遍歷字串，將每輪的得分壓入容器中。

#include <iostream>
#include <vector>
#include <cstdlib>
using namespace std;
class Solution {
public:
    int calPoints(vector<string>& ops) {
        //再建立一個容器
        vector<int> vec;
        for(int i = 0; i < ops.size(); i++){
            if(ops[i] == "C"){
                vec.pop_back();
            }
            else if(ops[i] == "D"){
                vec.push_back(vec[vec.size() - 1]*2);
            }
            else if(ops[i] == "+"){
                vec.push_back(vec[vec.size() - 1] + vec[vec.size() - 2]);
            }
            else{
                vec.push_back(atoi(ops[i].c_str()));
            }
        }

        int sum = 0;
        for(int j = 0; j < vec.size() ; j++){
            sum = sum + vec[j];
        }

        return sum;
    }
};
int main()
{
    Solution a;
    vector<string> s;
    s.push_back("5");
    s.push_back("2");
    s.push_back("C");
    s.push_back("D");
    s.push_back("+");
    cout << a.calPoints(s) << endl;
    return 0;
}

遇到的新知識：C++中如何將字串轉換為int型 參考文章

使用atoi(string) 可以將string型轉換成int型。在這裡還要先將字元型變成string型，使用 c_str()。

結果

執行用時: 4 ms, 在Baseball Game的C++提交中擊敗了99.56% 的使用者