682. Baseball Game-优快云博客

本文介绍了一种棒球比赛中的计分算法实现方法。针对输入的字符串列表，该算法能够处理四种不同类型的指令：整数分数、加号表示两轮分数之和、大写字母D表示上一轮分数的两倍以及大写字母C用于移除前一轮的有效分数。通过使用栈来存储每轮的得分，并在最后进行求和，从而得出整个比赛的总得分。

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You’re now a baseball game point recorder.

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

Integer (one round’s score): Directly represents the number of points you get in this round.
“+” (one round’s score): Represents that the points you get in this round are the sum of the last two valid round’s points.
“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.
“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.

思路:用一个栈,存下所有得分,最后求和即可.

class Solution {
public:
    int calPoints(vector<string>& ops) {
        stack<int> s;
        int sum = 0;
        for (int i = 0; i < ops.size(); i++) {//遍历ops,对每种情况处理
            if (ops[i] == "C") {
                s.pop();
            }
            else if (ops[i] == "D") {
                int tmp = s.top() * 2;
                s.push(tmp);
            }
            else if (ops[i] == "+") {
                int tmp1 = s.top();
                s.pop();
                int tmp2 = s.top() + tmp1;
                s.push(tmp1);
                s.push(tmp2);
            }
            else { // 若是数字
                int tmp = stoi(ops[i]);
                s.push(tmp);
            }
        }
        while(!s.empty()) {  //求和
            sum += s.top();
            s.pop();
        }
        return sum;
    }
};