[LeetCode] Peeking Iterator

[LeetCode] Peeking Iterator

Given an Iterator class interface with methods:next()AndhasNext(), Design and implement a PeekingIterator that supportspeek()Operation-it essential tially peek () at the element that will be returned by the next call to next ().

Here is an example. Assume that the iterator is initialized to the beginning of the list:[1, 2, 3].

Callnext()Gets you 1, the first element in the list.

Now you callpeek()And it returns 2, the next element. Callingnext()After that still return 2.

You callnext()The final time and it returns 3, the last element. CallinghasNext()After that shoshould return false.

Hint:

Think of "looking ahead". You want to cache the next element. Is one variable sufficient? Why or why not? Test your design with call order of peek () before next () vs next () before peek (). For a clean implementation, check out Google's guava library source code. Solutions

Use a variable to save the value of the peek element.next()Andpeek()Update the value.

Implementation Code

Java:

// Runtime: 120 ms//Java Iterator interface reference://https://docs.oracle.com/javase/8/docs/api/java/util/Iterator.htmlclass PeekingIterator implements Iterator
  
    {    Integer peek = null;    private Iterator
   
     iterator;    public PeekingIterator(Iterator
    
      iterator) {        this.iterator = iterator;    } // Returns the next element in the iteration without advancing the iterator.    public Integer peek() {        if (peek == null) {            peek = iterator.next();        }        return peek;    }    // hasNext() and next() should behave the same as in the Iterator interface.    // Override them if needed.    @Override    public Integer next() {        if (peek == null) {            return iterator.next();        }        else {            Integer temp = peek;            peek = null;            return temp;        }    }    @Override    public boolean hasNext() {        if (peek != null) {            return true;        }        else {            return iterator.hasNext();        }    }}