LeetCode 604 - Design Compressed String Iterator

http://bookshadow.com/weblog/2017/06/11/leetcode-design-compressed-string-iterator/

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 ' '

https://discuss.leetcode.com/topic/92098/java-concise-single-queue-solution

   Queue<int[]> queue = new LinkedList<>();
    
    public StringIterator(String s) {
        int i = 0, n = s.length();
        while (i < n) {
            int j = i+1;
            while (j < n && s.charAt(j) - 'A' < 0) j++;
            queue.add(new int[]{s.charAt(i) - 'A',  Integer.parseInt(s.substring(i+1, j))});
            i = j;
        }
    }
    
    public char next() {
        if (queue.isEmpty()) return ' ';
        int[] top = queue.peek();
        if (--top[1] == 0) queue.poll();
        return (char) ('A' + top[0]);
    }
    
    public boolean hasNext() {
        return !queue.isEmpty();
    }

https://www.nowtoshare.com/en/Article/Index/70578

public class StringIterator { Deque<Character> q; char cur; int count; public StringIterator(String compressedString) { q = new LinkedList<>(); cur ='#'; count=0; for(char ch : compressedString.toCharArray()){ q.offer(ch); } } public char next() { if(!hasNext()) return ' '; count--; return cur; } public boolean hasNext() { if(count>0) return true; else{ // System.out.println(q); if(q.size()==0) return false; else{ cur =q.pollFirst(); count=0; while(!q.isEmpty() && Character.isDigit(q.peekFirst())){ count*=10; count+=q.pollFirst()-'0'; } return true; } } } }

设计和实现压缩字符串的迭代器，支持函数：next和hasNext。

压缩字符串以字母+出现次数的格式给出。

将压缩字符串compressedString拆分成字母数组chars和出现次数数组times。

记原始字符串大小为size

变量idx记录chars的当前下标，cnt记录当前已经遍历过的字符个数

维护idx和cnt即可。

private ArrayList<Character> chars = new ArrayList<>(); private ArrayList<Long> times = new ArrayList<>(); private long size, cnt; private int idx; public StringIterator(String compressedString) { StringBuilder s = new StringBuilder(); for (char c : (compressedString + "#").toCharArray()) { if (c >= '0' && c <= '9') { s.append(c); continue; } if (s.length() > 0) { size += Integer.parseInt(s.toString()); times.add(this.size); s = new StringBuilder(); } if (c != '#') chars.add(c); } } public char next() { if (!hasNext()) return ' '; if (times.get(idx) < ++cnt) idx++; return chars.get(idx); } public boolean hasNext() { return cnt < size; }

X.
http://blog.csdn.net/kangbin825/article/details/73044743

1. public class StringIterator {  
2.   
3.     private String compressedString;  
4.     int c = 0;  
5.     private int pos = 0;  
6.     private char ch;  
7.   
8.     public StringIterator(String compressedString) {  
9.         this.compressedString = compressedString;  
10.     }  
11.   
12.     public char next() {  
13.         if (c > 0) {  
14.             c--;  
15.             return ch;  
16.         }  
17.   
18.         if (pos >= compressedString.length()) {  
19.             return ' ';  
20.         }  
21.   
22.         ch = compressedString.charAt(pos++);  
23.         while (pos < compressedString.length()  
24.                 && isDigit(compressedString.charAt(pos))) {  
25.             c = c * 10 + compressedString.charAt(pos++) - '0';  
26.         }  
27.         c--;  
28.         return ch;  
29.     }  
30.   
31.     private boolean isDigit(char charAt) {  
32.         return Character.isDigit(charAt);  
33.     }  
34.   
35.     public boolean hasNext() {  
36.         return c > 0 || pos < compressedString.length();  
37.     }  
38. }  

https://discuss.leetcode.com/topic/92107/java-solution-two-lists

public class StringIterator {
    List<Character> chars = new ArrayList<>();
    List<Integer> counts = new ArrayList<>();
    int ptr = 0;
    
    public StringIterator(String str) {
        int i = 0;
        while (i < str.length()) {
            chars.add(str.charAt(i));
            int j = i + 1;
            while (j < str.length() && Character.isDigit(str.charAt(j))) j++;
            counts.add(Integer.parseInt(str.substring(i + 1, j)));
            i = j;
        }
    }
    
    public char next() {
        if (!hasNext()) return ' ';
        
        char result = chars.get(ptr);
        counts.set(ptr, counts.get(ptr) - 1);
        if (counts.get(ptr) == 0) ptr++;
        return result;
    }
    
    public boolean hasNext() {
        return ptr < chars.size();
    }
}

https://discuss.leetcode.com/topic/92108/c-java-clean-code


https://discuss.leetcode.com/topic/92202/a-bit-of-java8-regex-and-very-short-solution/2

public class StringIterator {
    int i;
    String[] arr;
    int[] counts;

    public StringIterator(String str) {
        arr = str.split("\\d+");
        counts = Arrays.stream(str.substring(1).split("[a-zA-Z]+")).mapToInt(Integer::parseInt).toArray();
    }
    
    public char next() {
        if(!hasNext()) return ' ';
        char ch = arr[i].charAt(0);
        if(--counts[i] == 0) ++i;
        return ch;
    }

    public boolean hasNext() {
        if(i == arr.length) return false;
        return true;
    }
}