add prob #1209; O(N) in time and O(N) in space;

Author: randomwalk10

1209_Remove_All_Adjacent_Duplicates_in_String_II.cpp

@@ -0,0 +1,89 @@
+/*
+Given a string s, a k duplicate removal consists of choosing k adjacent and equal letters from s and removing them causing the left and the right side of the deleted substring to concatenate together.
+
+We repeatedly make k duplicate removals on s until we no longer can.
+
+Return the final string after all such duplicate removals have been made.
+
+It is guaranteed that the answer is unique.
+
+ 
+
+Example 1:
+
+Input: s = "abcd", k = 2
+Output: "abcd"
+Explanation: There's nothing to delete.
+Example 2:
+
+Input: s = "deeedbbcccbdaa", k = 3
+Output: "aa"
+Explanation: 
+First delete "eee" and "ccc", get "ddbbbdaa"
+Then delete "bbb", get "dddaa"
+Finally delete "ddd", get "aa"
+Example 3:
+
+Input: s = "pbbcggttciiippooaais", k = 2
+Output: "ps"
+ 
+
+Constraints:
+
+1 <= s.length <= 10^5
+2 <= k <= 10^4
+s only contains lower case English letters.
+
+来源：力扣（LeetCode）
+链接：https://leetcode-cn.com/problems/remove-all-adjacent-duplicates-in-string-ii
+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 <stack>
+using namespace std;
+
+class Solution {
+public:
+    string removeDuplicates(string s, int k) {
+        stack<char> a;
+        stack<int> b;
+        string res;
+
+        for(int i=0; i<(int)s.size(); ++i){
+            if( (!a.empty())&&(a.top()==s[i]) ){
+                if(b.top()==k-1){
+                    a.pop();
+                    b.pop();
+                }
+                else{
+                    int temp = b.top(); b.pop();
+                    b.push(temp+1);
+                }
+            }
+            else{
+                a.push(s[i]);
+                b.push(1);
+            }
+        }
+
+        stack<char> a_r;
+        stack<int> b_r;
+        while(!a.empty()){
+            a_r.push(a.top());
+            b_r.push(b.top());
+            a.pop();
+            b.pop();
+        }
+        while(!a_r.empty()){
+            for(int i=0; i<b_r.top(); ++i){
+                res.push_back(a_r.top());
+            }
+            a_r.pop();
+            b_r.pop();
+        }
+
+        return res;
+    }
+};