# 284-peeking-iterator


Try it on <a href='https://leetcode.com/problems/284-peeking-iterator'>leetcode</a>

## Description
<div class="description">
<div><p>Design an iterator that supports the <code>peek</code> operation on an existing iterator in addition to the <code>hasNext</code> and the <code>next</code> operations.</p>

<p>Implement the <code>PeekingIterator</code> class:</p>

<ul>
	<li><code>PeekingIterator(Iterator&lt;int&gt; nums)</code> Initializes the object with the given integer iterator <code>iterator</code>.</li>
	<li><code>int next()</code> Returns the next element in the array and moves the pointer to the next element.</li>
	<li><code>boolean hasNext()</code> Returns <code>true</code> if there are still elements in the array.</li>
	<li><code>int peek()</code> Returns the next element in the array <strong>without</strong> moving the pointer.</li>
</ul>

<p><strong>Note:</strong> Each language may have a different implementation of the constructor and <code>Iterator</code>, but they all support the <code>int next()</code> and <code>boolean hasNext()</code> functions.</p>

<p>&nbsp;</p>
<p><strong>Example 1:</strong></p>

<pre><strong>Input</strong>
["PeekingIterator", "next", "peek", "next", "next", "hasNext"]
[[[1, 2, 3]], [], [], [], [], []]
<strong>Output</strong>
[null, 1, 2, 2, 3, false]

<strong>Explanation</strong>
PeekingIterator peekingIterator = new PeekingIterator([1, 2, 3]); // [<u><strong>1</strong></u>,2,3]
peekingIterator.next();    // return 1, the pointer moves to the next element [1,<u><strong>2</strong></u>,3].
peekingIterator.peek();    // return 2, the pointer does not move [1,<u><strong>2</strong></u>,3].
peekingIterator.next();    // return 2, the pointer moves to the next element [1,2,<u><strong>3</strong></u>]
peekingIterator.next();    // return 3, the pointer moves to the next element [1,2,3]
peekingIterator.hasNext(); // return False
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= nums.length &lt;= 1000</code></li>
	<li><code>1 &lt;= nums[i] &lt;= 1000</code></li>
	<li>All the calls to <code>next</code> and <code>peek</code> are valid.</li>
	<li>At most <code>1000</code> calls will be made to <code>next</code>, <code>hasNext</code>, and <code>peek</code>.</li>
</ul>

<p>&nbsp;</p>
<strong>Follow up:</strong> How would you extend your design to be generic and work with all types, not just integer?</div>
</div>

## Solution(Python)
```Python
class PeekingIterator:
    def __init__(self, iterator):
        self.iterator = iterator
        self.buffer = self.iterator.next() if self.iterator.hasNext() else None

    def peek(self):
        return self.buffer

    def next(self):
        tmp = self.buffer
        self.buffer = self.iterator.next() if self.iterator.hasNext() else None
        return tmp

    def hasNext(self):
        return self.buffer != None

```