add solution of problem 394: decode string

Author: ppulse

DecodeString394/README.md

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+Given an encoded string, return it's decoded string.
+
+The encoding rule is: `k[encoded_string]`, where the *encoded_string* inside the square brackets is being repeated exactly *k* times. Note that *k* is guaranteed to be a positive integer.
+
+You may assume that the input string is always valid; No extra white spaces, square brackets are well-formed, etc.
+
+Furthermore, you may assume that the original data does not contain any digits and that digits are only for those repeat numbers, *k*. For example, there won't be input like `3a` or `2[4]`.
+
+#####Examples:
+```
+s = "3[a]2[bc]", return "aaabcbc".
+s = "3[a2[c]]", return "accaccacc".
+s = "2[abc]3[cd]ef", return "abcabccdcdcdef".
+```

DecodeString394/class Solution.java

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+public class Solution {
+    public String decodeString(String s) {
+        if (s == null)
+            return "";
+        
+        int sLen = s.length();
+        int idx = 0;
+        StringBuilder sb = new StringBuilder();
+        
+        while (idx < sLen) {
+        	int idxOfFirstLetter = idx;
+            while (idx < sLen && !(s.charAt(idx) >= '0' && s.charAt(idx) <= '9')) {
+                idx++;
+            }
+        
+            int idxOfFirstDigit = idx;
+            while (idx < sLen && s.charAt(idx) != '[') {
+                idx++;
+            }
+        
+            sb.append(s.substring(idxOfFirstLetter, idxOfFirstDigit));
+            if (idx < sLen) { 
+                int factor = Integer.valueOf(s.substring(idxOfFirstDigit, idx));
+        
+                idx++;
+        
+                int startIdx = idx;
+                int cnt = 0;
+                while (idx < sLen && (s.charAt(idx) != ']' || cnt > 0)) {
+                    if (s.charAt(idx) == '[')
+                        cnt++;
+                    if (s.charAt(idx) == ']')
+                        cnt--;
+                    idx++;
+                }      
+        
+                String subStr = decodeString(s.substring(startIdx, idx));    
+            
+                for (int i = 0; i < factor; i++) {
+                    sb.append(subStr);
+                }
+            
+                idx++;
+            }
+        }
+        
+        return sb.toString();
+    }
+}