Baseball Game

https://leetcode.com/problems/baseball-game/description/

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.

Thoughts

要把上次或上两次存入的数拿出来，栈的一个常见应用。

Code

class Solution {
    private boolean isNum(String str) {
        if (str.length() == 0) {
            return false;
        }
        if ("-0123456789".contains("" + str.charAt(0))) {
            return true;
        }
        return false;
    }

    public int calPoints(String[] ops) {
        Stack<Integer> scores = new Stack<>();
        for (String str : ops) {
            if (isNum(str)) {
                scores.push(Integer.parseInt(str));
            } else {
                switch (str) {
                    case "C":
                        if (!scores.isEmpty()) {
                            scores.pop();
                        }
                        break;
                    case "D":
                        if (!scores.isEmpty()) {
                            scores.push(scores.peek() * 2);
                        }
                        break;
                    case "+":
                        if (scores.size() >= 2) {
                            int first = scores.pop();
                            int sec = scores.pop();
                            scores.push(sec);
                            scores.push(first);
                            scores.push(first + sec);
                        }
                        break;
                    default:
                        break;
                }
            }
        }

        int res = 0;
        for (Integer num : scores) {
            res += num;
        }

        return res;
    }
}

Analysis

做题耗时15min

时空复杂度O(n).