Baseball Game: A Stack Solution
You can find the problem here (#422 in my solved list):
https://leetcode.com/problems/baseball-game/
The size of the input list will be between 1 and 1000.
Every integer represented in the list will be between -30000 and 30000.
This problem screams for a stack solution. As you go thru the operations array, simply follow the rules as stipulated in the problem description, by pushing and popping to/from a stack accordingly. Just be careful to push back into the stack whatever you pop out, in some cases.
Straightforward solution with a reasonable execution time. Cheers, ACC.
https://leetcode.com/problems/baseball-game/
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 twovalidround's points."D"(one round's score): Represents that the points you get in this round are the doubled data of the lastvalidround's points."C"(an operation, which isn't a round's score): Represents the lastvalidround'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:
This problem screams for a stack solution. As you go thru the operations array, simply follow the rules as stipulated in the problem description, by pushing and popping to/from a stack accordingly. Just be careful to push back into the stack whatever you pop out, in some cases.
Straightforward solution with a reasonable execution time. Cheers, ACC.
public class Solution
{
public int CalPoints(string[] ops)
{
Stack stack = new Stack();
for (int i = 0; i < ops.Length; i++)
{
int val = 0;
if (Int32.TryParse(ops[i], out val))
{
stack.Push(val);
}
else if (ops[i].Equals("+"))
{
int v1 = stack.Pop();
int v2 = stack.Pop();
stack.Push(v2);
stack.Push(v1);
stack.Push(v1 + v2);
}
else if (ops[i].Equals("D"))
{
int v1 = stack.Pop();
stack.Push(v1);
stack.Push(2 * v1);
}
else if (ops[i].Equals("C"))
{
stack.Pop();
}
}
int result = 0;
while (stack.Count > 0)
{
result += stack.Pop();
}
return result;
}
}
Comments
I love stacks!
Reply Deleteclass Solution {
public:
int calPoints(vector& ops) {
vector operands; operands.reserve(ops.size());
for (const auto& op : ops) {
if (op[0] == '-' || isdigit(op[0])) {
operands.push_back(stoi(op));
} else if (op == "+") {
operands.push_back(operands.back() + operands[operands.size()-2]);
} else if (op == "D") {
operands.push_back(operands.back() * 2);
} else if (op == "C") {
operands.pop_back();
}
}
return accumulate(operands.cbegin(), operands.cend(), 0);
}
};
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