add prob #682; O(N) in time and O(N) in space;

Author: randomwalk10

682_Baseball_Game.cpp

@@ -0,0 +1,89 @@
+/*
+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.
+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.
+
+来源：力扣（LeetCode）
+链接：https://leetcode-cn.com/problems/baseball-game
+著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。
+*/
+
+/* O(N) in time and O(N) in space */
+
+#include <string>
+#include <vector>
+#include <stack>
+using namespace std;
+
+class Solution {
+public:
+    int calPoints(vector<string>& ops) {
+        stack<int> s;
+        int res = 0;
+
+        for(int i=0; i<(int)ops.size(); ++i){
+            if(ops[i][0]=='+'){
+                int prev1 = s.top(); s.pop();
+                int prev2 = s.top(); s.pop();
+                s.push(prev2);
+                s.push(prev1);
+                s.push(prev1+prev2);
+            }
+            else if(ops[i][0]=='D'){
+                s.push(2*s.top());
+            }
+            else if(ops[i][0]=='C'){
+                s.pop();
+            }
+            else{
+                int temp = 0;
+                bool is_pos = true;
+                for(int j=0; j<(int)ops[i].size(); ++j){
+                    if(ops[i][j]=='-') is_pos = false;
+                    else temp = temp*10 + ops[i][j] - '0';
+                }
+                if(is_pos) s.push(temp);
+                else s.push(-temp);
+            }
+        }
+
+        while(!s.empty()){
+            res += s.top(); s.pop();
+        }
+
+        return res;
+    }
+};