LeetCode 604. Design Compressed String Iterator (设计压缩字符迭代器)$
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.
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 ' '
题目标签:Design
Java Solution:
Runtime beats 86.01%
完成日期:07/10/2017
关键词:Design
关键点:用两个list和一个cursor来找到char和对应char的count
public class StringIterator
{
ArrayList<Integer> counts;
ArrayList<Character> letters;
int cursor_index = 0; public StringIterator(String compressedString)
{
counts = new ArrayList<>();
letters = new ArrayList<>(); for(int i=0; i<compressedString.length(); i++)
{
// if find a letter
if(Character.isLetter(compressedString.charAt(i)))
letters.add(compressedString.charAt(i)); // if find a digit
else if(Character.isDigit(compressedString.charAt(i)))
{
int end = i+1;
// find the next non-digit char within the length
while(end < compressedString.length() &&
Character.isDigit(compressedString.charAt(end)))
end++; counts.add(Integer.parseInt(compressedString.substring(i, end))); i = end - 1;
}
}
} public char next()
{
char c; // meaning string is finished
if(!hasNext())
return ' '; c = letters.get(cursor_index);
counts.set(cursor_index, counts.get(cursor_index) - 1); if(counts.get(cursor_index) == 0)
cursor_index++; return c;
} public boolean hasNext()
{
// if string is finished
if(cursor_index > counts.size() - 1 )
return false;
else // if string is not finished
return true;
}
} /**
* Your StringIterator object will be instantiated and called as such:
* StringIterator obj = new StringIterator(compressedString);
* char param_1 = obj.next();
* boolean param_2 = obj.hasNext();
*/
参考资料:N/A
LeetCode 算法题目列表 - LeetCode Algorithms Questions List
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