LeetCode Design Compressed String Iterator

最新推荐文章于 2025-09-07 20:32:33 发布

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最新推荐文章于 2025-09-07 20:32:33 发布

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文章标签：数据结构与算法 java

原文链接：http://www.cnblogs.com/Dylan-Java-NYC/p/7555412.html

本文介绍了一种用于处理压缩字符串的迭代器设计方案。该方案通过实现next和hasNext方法，支持对压缩字符串进行高效遍历。文章详细解释了算法的工作原理，并提供了完整的Java代码示例。

原题链接在这里：https://leetcode.com/problems/design-compressed-string-iterator/description/

题目：

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 ' '

题解：

用 i 标记当前index位置. 用count计数之前char出现的次数. 当count--到0时继续向后移动 i.

Time Complexity: hasNext(), O(1). next(), O(n). n为数字串的长度.

Space: O(1).

AC Java:

 1 class StringIterator {
 2     String s;
 3     int i = 0;
 4     int count = 0;
 5     
 6     char res;
 7     public StringIterator(String compressedString) {
 8         this.s = compressedString;    
 9     }
10     
11     public char next() {
12         if(!hasNext()){
13             return ' ';
14         }
15 
16         if(count == 0){
17             res = s.charAt(i);
18             i++;
19             while(i<s.length() && Character.isDigit(s.charAt(i))){
20                 //Overflow
21                 if(count > Integer.MAX_VALUE/10 || (count == Integer.MAX_VALUE/10 && s.charAt(i)>='8')){
22                     throw new IllegalArgumentException("Repeated number is larger than Integer.MAX_VALUE.");
23                 }
24                 
25                 count = count*10 + (s.charAt(i)-'0');
26                 i++;
27             }
28         }
29         count--;
30         return res;
31     }
32     
33     public boolean hasNext() {
34         return !(i==s.length() && count==0);
35     }
36 }
37 
38 /**
39  * Your StringIterator object will be instantiated and called as such:
40  * StringIterator obj = new StringIterator(compressedString);
41  * char param_1 = obj.next();
42  * boolean param_2 = obj.hasNext();
43  */