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.

解法:栈,用一个stack记录,循环字符串,如果是数字直接进栈,如果当前字符是操作符对栈里面的数字进行相应处理,'+'把栈里面最后两个数字相加然后如栈,'D'把栈里的最后一个数字乘以2然后如栈,'C'移除栈里面的最后一个元素。最后返回栈里所有元素的和。

Java:

class Solution {
public int calPoints(String[] ops) {
int sum = 0;
LinkedList<Integer> list = new LinkedList<>();
for (String op : ops) {
if (op.equals("C")) {
sum -= list.removeLast();
}
else if (op.equals("D")) {
list.add(list.peekLast() * 2);
sum += list.peekLast();
}
else if (op.equals("+")) {
list.add(list.peekLast() + list.get(list.size() - 2));
sum += list.peekLast();
}
else {
list.add(Integer.parseInt(op));
sum += list.peekLast();
}
}
return sum;
}
}  

Python:

class Solution(object):
def calPoints(self, ops):
"""
:type ops: List[str]
:rtype: int
"""
history = []
for op in ops:
if op == '+':
history.append(history[-1] + history[-2])
elif op == 'D':
history.append(history[-1] * 2)
elif op == 'C':
history.pop()
else:
history.append(int(op))
return sum(history)  

C++:

class Solution {
public:
int calPoints(vector<string>& ops) {
vector<int> v;
for (string op : ops) {
if (op == "+") {
v.push_back(v.back() + v[v.size() - 2]);
} else if (op == "D") {
v.push_back(2 * v.back());
} else if (op == "C") {
v.pop_back();
} else {
v.push_back(stoi(op));
}
}
return accumulate(v.begin(), v.end(), 0);
}
};

  

All LeetCode Questions List 题目汇总

最新文章

  1. Month Scheme
  2. Nodejs 创建web服务
  3. [ASP.NET MVC 小牛之路]13 - Helper Method
  4. Hadoop JAVA 开发说明
  5. 项目管理详细任务(PMBOK2008)
  6. linq group by多个字段,返回多个字段.
  7. DeviceOne接受了DevStore的专访
  8. 在WPF应用程序中利用IEditableObject接口实现可撤销编辑的对象
  9. ZeroClipboard实现复制
  10. Java用通配符 获得泛型的协变和逆变
  11. Overcome the Dilemma of &quot;unlock&quot; and &quot;trust&quot;
  12. python Django 学习笔记(三)—— 模版的使用
  13. Datatable转换成List实体对象列表 几个实例
  14. javascript笔记之正则表达式
  15. HDOJ(HDU) 1785 You Are All Excellent(角度运算)
  16. iOS实现白板、画板功能,有趣的涂鸦工具,已封装,简单快捷使用
  17. 三种读取HashMap的方式
  18. Windows内核中的内存管理
  19. ionic2项目中实现md5加密
  20. ansible笔记(6):常用模块之命令类模块

热门文章

  1. 微信小程序~跳页传参
  2. postgres高可用学习篇二:通过pgbouncer连接池工具来管理postgres连接
  3. 跨子域的iframe高度自适应
  4. 【JZOJ6226】【20190618】纳什均衡
  5. NuxtJS如何利用axios异步请求
  6. leetcode题目讲解(Python):字符串转整数 (atoi)
  7. 咕泡学院java架构vip课程
  8. 通过phoenix导入数据到hbase出错记录
  9. mysql5.7 之 sql_mode=only_full_group_by问题
  10. 【洛谷】P4198 楼房重建(线段树)