604. Design Compressed String Iterator π
Description
Design and implement a data structure for a compressed string iterator. The given compressed string will be in the form of each letter followed by a positive integer representing the number of this letter existing in the original uncompressed string.
Implement the StringIterator class:
next()
Returns the next character if the original string still has uncompressed characters, otherwise returns a white space.hasNext()
Returns true if there is any letter needs to be uncompressed in the original string, otherwise returnsfalse
.
Example 1:
Input ["StringIterator", "next", "next", "next", "next", "next", "next", "hasNext", "next", "hasNext"] [["L1e2t1C1o1d1e1"], [], [], [], [], [], [], [], [], []] Output [null, "L", "e", "e", "t", "C", "o", true, "d", true] Explanation StringIterator stringIterator = new StringIterator("L1e2t1C1o1d1e1"); stringIterator.next(); // return "L" stringIterator.next(); // return "e" stringIterator.next(); // return "e" stringIterator.next(); // return "t" stringIterator.next(); // return "C" stringIterator.next(); // return "o" stringIterator.hasNext(); // return True stringIterator.next(); // return "d" stringIterator.hasNext(); // return True
Constraints:
1 <= compressedString.length <= 1000
compressedString
consists of lower-case an upper-case English letters and digits.- The number of a single character repetitions in
compressedString
is in the range[1, 10^9]
- At most
100
calls will be made tonext
andhasNext
.
Solutions
Solution 1
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1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 |
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1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 |
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