LeetCode 604. Design Compressed String Iterator(java)
2018-02-07 08:19
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Design and implement a data structure for a compressed string iterator. It should support the following operations: next and hasNext.
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.
next() - if the original string still has uncompressed characters, return the next letter; Otherwise return a white space.
hasNext() - Judge whether there is any letter needs to be uncompressed.
Note:
Please remember to RESET your class variables declared in StringIterator, as static/class variables are persisted across multiple test cases. Please see here for more details.
思路:首先,我想的是用一个stack,从后往前,把最后生成的字符都塞进stack,然后每次next()和hasNext()的时候,就从stack中pop出来。但是这个方法过不了leetcode,因为出现了这样一个例子“a1038373845b1032394854”,导致时间非常慢,所以我又换了一个方法,从左边开始,每次pop都计数-1,当计数为0了,再开始第二个字符。以下为两个方法的代码,但只有第二个方法可以过LeetCode。
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.
next() - if the original string still has uncompressed characters, return the next letter; Otherwise return a white space.
hasNext() - Judge whether there is any letter needs to be uncompressed.
Note:
Please remember to RESET your class variables declared in StringIterator, as static/class variables are persisted across multiple test cases. Please see here for more details.
Example: StringIterator iterator = new StringIterator("L1e2t1C1o1d1e1"); iterator.next(); // return 'L' iterator.next(); // return 'e' iterator.next(); // return 'e' iterator.next(); // return 't' iterator.next(); // return 'C' iterator.next(); // return 'o' iterator.next(); // return 'd' iterator.hasNext(); // return true iterator.next(); // return 'e' iterator.hasNext(); // return false iterator.next(); // return ' '
思路:首先,我想的是用一个stack,从后往前,把最后生成的字符都塞进stack,然后每次next()和hasNext()的时候,就从stack中pop出来。但是这个方法过不了leetcode,因为出现了这样一个例子“a1038373845b1032394854”,导致时间非常慢,所以我又换了一个方法,从左边开始,每次pop都计数-1,当计数为0了,再开始第二个字符。以下为两个方法的代码,但只有第二个方法可以过LeetCode。
//代码一:stack class StringIterator { Stack<Character> stack = new Stack<>(); public StringIterator(String compressedString) { int count = 1, num = 0, len = compressedString.length(); for (int i = len - 1; i >= 0; i--) { char c = compressedString.charAt(i); if (Character.isDigit(c)) { int end = i; while (i < len && Character.isDigit(compressedString.charAt(i))) { i--; } i++; for (int j = i; j <= end; j++) { num = num * 10 + compressedString.charAt(j) - '0'; if (num > Integer.MAX_VALUE) { num = Integer.MAX_VALUE; } } count = num; num = 0; } else { for (int j = 0; j < count; j++) { stack.push(c); } count = 1; } } } public char next() { return stack.isEmpty() ? ' ' : stack.pop(); } public boolean hasNext() { return stack.isEmpty() ? false : true; } }
//代码二: class StringIterator { char letter; int count; int index; int len; String s; public StringIterator(String compressedString) { s = compressedString; len = s.length(); index = 0; helper(); } public char next() { if (count > 0) { count--; return letter; } else { if (index == len) { return ' '; } else { helper(); count--; return letter; } } } public boolean hasNext() { return (index == len && count <= 0) ? false : true; } public void helper() { char c = s.charAt(index); letter = c; int num = 0; while (++index < len && Character.isDigit(s.charAt(index))) { num = num * 10 + s.charAt(index) - '0'; } count = num; } }
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