【刷穿 LeetCode】284. 顶端迭代器 : 迭代器基本认识的模拟题

输入：
["PeekingIterator", "next", "peek", "next", "next", "hasNext"]
[[[1, 2, 3]], [], [], [], [], []]
输出：
[null, 1, 2, 2, 3, false]
解释：
PeekingIterator peekingIterator = new PeekingIterator([1, 2, 3]); // [1,2,3]
peekingIterator.next();    // 返回 1 ，指针移动到下一个元素 [1,2,3]
peekingIterator.peek();    // 返回 2 ，指针未发生移动 [1,2,3]
peekingIterator.next();    // 返回 2 ，指针移动到下一个元素 [1,2,3]
peekingIterator.next();    // 返回 3 ，指针移动到下一个元素 [1,2,3]
peekingIterator.hasNext(); // 返回 False
复制代码

class PeekingIterator implements Iterator<Integer> {
    Iterator<Integer> iter;
    Integer next;
    public PeekingIterator(Iterator<Integer> iterator) {
          iter = iterator;
          if (iter.hasNext()) next = iter.next();
    }
    public Integer peek() {
          return next;
    }
    @Override
    public Integer next() {
          Integer ans = next;
          next = iter.hasNext() ? iter.next() : null;
        return ans;
    }
    @Override
    public boolean hasNext() {
          return next != null;
    }
}
复制代码

class PeekingIterator implements Iterator<E> {
    Iterator<E> iter;
    E next;
    public PeekingIterator(Iterator<E> iterator) {
          iter = iterator;
          if (iter.hasNext()) next = iter.next();
    }
    public E peek() {
          return next;
    }
    @Override
    public E next() {
          E ans = next;
          next = iter.hasNext() ? iter.next() : null;
        return ans;
    }
    @Override
    public boolean hasNext() {
          return next != null;
    }
  }
复制代码