284. Peeking Iterator

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题目：

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:

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

Follow up: How would you extend your design to be generic and work with all types, not just integer?

链接： http://leetcode.com/problems/peeking-iterator/

题解：

设计一个带peek()的Iterator。这里我们根据题意和提示，提前cached下一个元素就可以了。这里我用了一个boolean变量 peekUsed来判断我们是否使用过peek()，其实也可以不用。在用过peek()之后，我们只返回cache过的element，而且hasNext() = true。否则我们使用正常的iterator的methods。

Time Complexity - O(1)， Space Complexity - O(1)

// Java Iterator interface reference:
// https://docs.oracle.com/javase/8/docs/api/java/util/Iterator.html
class PeekingIterator implements Iterator<Integer> {
    private Iterator<Integer> iter;
    private Integer nextElement;
    private boolean peekUsed;
    
    public PeekingIterator(Iterator<Integer> iterator) {
        // initialize any member here.
        iter = iterator;
    }

    // Returns the next element in the iteration without advancing the iterator.
    public Integer peek() {
        if(!peekUsed) {
            nextElement = iter.next();
            peekUsed = true;
        }    
        return nextElement;
    }

    // hasNext() and next() should behave the same as in the Iterator interface.
    // Override them if needed.
    @Override
    public Integer next() {
        if(peekUsed) {
            peekUsed = false;
            return nextElement;
        }
        return iter.next();
    }

    @Override
    public boolean hasNext() {
        if(peekUsed) {
            return true;
        }
        return iter.hasNext();
    }
}

Reference:

https://leetcode.com/discuss/59368/concise-java-solution

https://leetcode.com/discuss/59327/my-java-solution

https://leetcode.com/discuss/62829/simple-java-solution-by-caching-next-element

284. Peeking Iterator

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原文地址：http://www.cnblogs.com/yrbbest/p/5037886.html