Baseball Game(leetcode)

Baseball Game

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题目

leetcode题目

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.

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解析

在给定的字符串组中，每个字符串代表每一轮的得分。

1. 当字符串为Integer的时候，代表了这轮得分是Integer。
2. 当字符串为+的时候，这轮得分是前两轮有效得分的总和。
3. 当字符串为D的时候，这轮得分是上一轮有效得分的2倍。
4. 当字符串为C的时候，上轮得分无效。

因此，我们需要一个容器来记录有效得分，以方便我们计算最后的得分。

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解决

1. stack解决

利用stack容器将有效得分记录下来，然后再求stack中数的总和。

class Solution {
public:
    int calPoints(vector<string>& ops) {
        stack<int> points;
        int sum = 0;
        if (ops.size() == 0) {
            return sum;
        }
        int num = ops.size();
        for (int i = 0; i < num; i++) {
            int temp = 0;
            if (ops[i] == "C") {
                points.pop();
            } else if (ops[i] == "D") {
                temp = points.top() * 2;
                points.push(temp);
            } else if (ops[i] == "+") {
                int a = points.top();
                points.pop();
                int b = points.top();
                points.push(a);
                points.push(a + b);
            } else {
                temp = stoi(ops[i]);
                points.push(temp);
            }
        }
        while (!points.empty()) {
            sum += points.top();
            points.pop();
        }
        return sum;
    }
};

2. vector解决

class Solution {
public:
    int calPoints(vector<string>& ops) {
        vector<int> points;
        if (ops.size() == 0) {
            return 0;
        }
        int sum = 0;
        int num = ops.size();
        for (int i = 0; i < num; i++) {
            if (ops[i] == "C") {
                sum -= points[points.size() - 1];
                points.pop_back();
            } else if (ops[i] == "D") {
                sum += points[points.size() - 1] * 2;
                points.push_back(points[points.size() - 1] * 2);
            } else if (ops[i] == "+") {
                sum += points[points.size() - 1] + points[points.size() - 2];
                points.push_back(points[points.size() - 1] + points[points.size() - 2]);
            } else {
                sum += stoi(ops[i]);
                points.push_back(stoi(ops[i]));
            }
        }
        return sum;
    }
};