Recursive

Author: Adam Lin

394_decodeString.cpp

@@ -0,0 +1,98 @@
+//https://www.youtube.com/watch?v=qB0zZpBJlh8
+
+class Solution {
+public:
+    string decodeString(string s) {
+        stack<string> stk; // Initialize a stack to store characters, numbers, and intermediate results
+
+        for(int i = 0; i < s.size(); ++i){
+            if(s[i] != ']'){
+                // If the current character is not ']', push it onto the stack
+                stk.push(string(1, s[i]));
+            } else {
+                // When we encounter a ']', we need to decode the substring inside the brackets
+                string substr = "";
+                // Pop characters until we find the matching '['
+                while(stk.top() != "["){
+                    substr = stk.top() + substr; // Build the substring in correct order
+                    stk.pop();
+                }
+                stk.pop(); // Remove the '[' from the stack
+
+                // Now, we need to find the number (could be more than one digit) before the '['
+                string k = "";
+                while(!stk.empty() && isdigit(stk.top()[0])){
+                    k = stk.top() + k; // Build the number in correct order
+                    stk.pop();
+                }
+                int num = stoi(k); // Convert the string number to an integer
+
+                // Repeat the substring 'num' times
+                string repeatedStr = "";
+                for(int j = 0; j < num; ++j){
+                    repeatedStr += substr;
+                }
+                // Push the repeated substring back onto the stack
+                stk.push(repeatedStr);
+            }
+        }
+
+        // Build the final result from the stack
+        string res = "";
+        while(!stk.empty()){
+            res = stk.top() + res; // Build the result in correct order
+            stk.pop();
+        }
+        return res; // Return the decoded string
+    }
+};
+
+// Recursive solution 
+class Solution {
+public:
+    int i = 0; // Class member to keep track of index
+
+    string decode(string& s) {
+        string result = "";
+        int num = 0;
+        while (i < s.size()) {
+            char c = s[i];
+            if (isdigit(c)) {
+                num = num * 10 + (c - '0');
+                i++;
+            } else if (c == '[') {
+                i++; // Move past '['
+                string decoded = decode(s);
+                // Repeat and append
+                while (num > 0) {
+                    result += decoded;
+                    num--;
+                }
+                num = 0;
+            } else if (c == ']') {
+                i++; // Move past ']'
+                return result;
+            } else {
+                result += c;
+                i++;
+            }
+        }
+        return result;
+    }
+
+    string decodeString(string s) {
+        i = 0; // Reset index before starting
+        return decode(s);
+    }
+};
+/*
+decode("3[a2[c]]", 0)
+|
++-- decode("3[a2[c]]", 2)
+    |
+    +-- decode("3[a2[c]]", 5)
+    |   - 返回 ("c", 7)
+    - 返回 ("acc", 8)
+- 返回 ("accaccacc", 8)
+
+*/