[leetcode] Peeking Iterator

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.

分析:利用一个Integer变量val保存peek()的结果,每次判断val是否为null,若不为null,则调用缓存的结果,否则调用Iterator<Integer>的相应的方法。

Java代码:

public class PeekingIterator implements Iterator<Integer> {
    Integer val;
    Iterator<Integer> iter;

    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 (val != null) {
            return val;
        }
        val = iter.next();
        return val;
    }

    // hasNext() and next() should behave the same as in the Iterator interface.
    // Override them if needed.
    @Override
    public Integer next() {
        if (val != null) {
            int tmp = val;
            val = null;
            return tmp;
        }
        return iter.next();
    }

    @Override
    public boolean hasNext() {
        if (val != null) {
            return true;
        }
        return iter.hasNext();
    }
}
原文地址:https://www.cnblogs.com/lasclocker/p/4938543.html