【LeetCode】284.Peeking Iterator

浅浅的花香味﹌ 2022-04-18 04:20 74阅读 0赞

### 284.Peeking Iterator ###

题目描述：

Given an Iterator class interface with methods: `next()` and `hasNext()`, design and implement a PeekingIterator that support the `peek()` operation – it essentially peek() at the element that will be returned by the next call to next().

**Example:**

Assume that the iterator is initialized to the beginning of the list: [1,2,3].
    
    Call next() gets you 1, the first element in the list.
    Now you call peek() and it returns 2, the next element. Calling next() after that still return 2. 
    You call next() the final time and it returns 3, the last element. 
    Calling hasNext() after that should return false.

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### 解析 ###

题目要求传入一个迭代器，然后实现`peek`功能，有点类似`Stack`与`Queue`的操作，在这两者中都存在`peek`的功能，即返回顶部元素，而不移除数据结构中的元素。由于队列的是`先进先出`，跟这个迭代器的迭代规则有点类似，因此我们可以选择使用`Queue`来模拟，初始化时，将迭代器中的元素填充到队列中，然后直接调用`Queue`的方法即可。

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### 代码 ###

// Java Iterator interface reference:
    // https://docs.oracle.com/javase/8/docs/api/java/util/Iterator.html
    class PeekingIterator implements Iterator<Integer> { 
        
        
        
        private Queue<Integer> queue = new LinkedList<>();;
    	public PeekingIterator(Iterator<Integer> iterator) { 
    	    // initialize any member here.
    	    while(iterator.hasNext()){ 
                queue.add(iterator.next());
            }
    	}
    
        // Returns the next element in the iteration without advancing the iterator.
    	public Integer peek() { 
            return queue.peek();
    	}
    
    	// hasNext() and next() should behave the same as in the Iterator interface.
    	// Override them if needed.
    	@Override
    	public Integer next() { 
    	    return queue.poll();
    	}
    
    	@Override
    	public boolean hasNext() { 
    	    return queue.size()>0;
    	}
    }