[LeetCode] Peeking Iterator

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().

Here is an 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.

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.

解题思路

用一个变量来保存peek元素值，每次调用

next()

和

peek()

时更新其值。

实现代码

Java：

// Runtime: 120 ms
//Java Iterator interface reference:
//https://docs.oracle.com/javase/8/docs/api/java/util/Iterator.html
class PeekingIterator implements Iterator<Integer> {
Integer peek = null;
private Iterator<Integer> iterator;
public PeekingIterator(Iterator<Integer> 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();
}
}
}