Design Compressed String Iterator (Easy Google)

update May 12,2018 16:24

LeetCode

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 ' '

Basic Idea:

设计一个可以读取压缩字符串的 iterator,和 decompression 不同的地方就在于这里需要用 class variable 保存当前的状态。

  • CPP Code:

      class StringIterator {
          string str;
          char chr;
          int remain;
          int currIndex;
    
          int getInt(int start) {
              int i = start;
              int num = 0;
              while (i < str.size() && str[i] >= '0' && str[i] <= '9') {
                  num = num * 10 + str[i] - '0';
                  i++;
              }
              currIndex = i;
              return num;
          }
      public:
          StringIterator(string compressedString) {
              this->str = compressedString;
              chr = 0;
              remain = 0;
              currIndex = 0;
          }
    
          char next() {
              if (! hasNext()) return ' ';
              --remain;
              return chr;
          }
    
          bool hasNext() {
              if (remain > 0) return true;
              else if (currIndex == str.size()) return false;
              chr = str[currIndex];
              remain = getInt(currIndex + 1);
              return true;
          }
      };
    
  • Java Code:

      class StringIterator {
          private String str;
          private int currIndex;
          private int remain;
          private char chr;
    
          public StringIterator(String compressedString) {
              this.str = compressedString;
              this.currIndex = 0;
              this.remain = 0;
              this.chr = 0;
          }
    
          public char next() {
              if (! hasNext()) return ' ';
              remain--;
              return chr;
          }
    
          public boolean hasNext() {
              if (remain > 0) return true;
              else if (remain == 0 && currIndex == str.length()) return false;
              chr = str.charAt(currIndex);
              remain = getInt(currIndex + 1);
              return true;
          }
    
          private int getInt(int start) {
              int num = 0;
              while (start < str.length() && str.charAt(start) >= '0' && str.charAt(start) <= '9') {
                  num = num * 10 + str.charAt(start) - '0';
                  start++;
              }
              currIndex = start;
              return num;
          }
      }
    

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