Medium
Given an encoded string, return its 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; there are 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 will not be input like 3a
or 2[4]
.
The test cases are generated so that the length of the output will never exceed 105
.
Example 1:
Input: s = “3[a]2[bc]”
Output: “aaabcbc”
Example 2:
Input: s = “3[a2[c]]”
Output: “accaccacc”
Example 3:
Input: s = “2[abc]3[cd]ef”
Output: “abcabccdcdcdef”
Constraints:
1 <= s.length <= 30
s
consists of lowercase English letters, digits, and square brackets '[]'
.s
is guaranteed to be a valid input.s
are in the range [1, 300]
.class Solution {
int i = 0;
String decodeString(String s) {
int count = 0;
StringBuffer sb = StringBuffer();
while (i < s.length) {
String c = s[i];
i++;
if (RegExp(r'[a-zA-Z]').hasMatch(c)) {
sb.write(c);
} else if (RegExp(r'\d').hasMatch(c)) {
count = count * 10 + int.parse(c);
} else if (c == ']') {
break;
} else if (c == '[') {
// Sub-problem: recursively decode the substring
String repeat = decodeString(s);
while (count > 0) {
sb.write(repeat);
count--;
}
}
}
return sb.toString();
}
}