顶端迭代器

284. 顶端迭代器

请你在设计一个迭代器，在集成现有迭代器拥有的 hasNext 和 next 操作的基础上，还额外支持 peek 操作。

实现 PeekingIterator 类：

PeekingIterator(Iterator<int> nums) 使用指定整数迭代器 nums 初始化迭代器。
int next() 返回数组中的下一个元素，并将指针移动到下个元素处。
bool hasNext() 如果数组中存在下一个元素，返回 true ；否则，返回 false 。
int peek() 返回数组中的下一个元素，但不移动指针。

注意：每种语言可能有不同的构造函数和迭代器 Iterator，但均支持 int next() 和 boolean hasNext() 函数。

示例 1：

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

提示：

1 <= nums.length <= 1000
1 <= nums[i] <= 1000
对 next 和 peek 的调用均有效
next、hasNext 和 peek 最多调用 1000 次

进阶：你将如何拓展你的设计？使之变得通用化，从而适应所有的类型，而不只是整数型？

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原站题解

上次编辑到这里，代码来自缓存点击恢复默认模板

/* * Below is the interface for Iterator, which is already defined for you. * **DO NOT** modify the interface for Iterator. * * class Iterator { * struct Data; * Data* data; * public: * Iterator(const vector<int>& nums); * Iterator(const Iterator& iter); * * // Returns the next element in the iteration. * int next(); * * // Returns true if the iteration has more elements. * bool hasNext() const; * }; */ class PeekingIterator : public Iterator { public: PeekingIterator(const vector<int>& nums) : Iterator(nums) { // Initialize any member here. // **DO NOT** save a copy of nums and manipulate it directly. // You should only use the Iterator interface methods. } // Returns the next element in the iteration without advancing the iterator. int peek() { } // hasNext() and next() should behave the same as in the Iterator interface. // Override them if needed. int next() { } bool hasNext() const { } };

java 解法, 执行用时: 4 ms, 内存消耗: 41.4 MB, 提交时间: 2022-11-12 12:05:51

// Java Iterator interface reference:
// https://docs.oracle.com/javase/8/docs/api/java/util/Iterator.html

class PeekingIterator implements Iterator<Integer> {
    private Iterator<Integer> iterator;
    private Integer nextElement;

    public PeekingIterator(Iterator<Integer> iterator) {
        this.iterator = iterator;
        nextElement = iterator.next();
    }
    
    public Integer peek() {
        return nextElement;
    }
    
    @Override
    public Integer next() {
        Integer ret = nextElement;
        nextElement = iterator.hasNext() ? iterator.next() : null;
        return ret;
    }
    
    @Override
    public boolean hasNext() {
        return nextElement != null;
    }
}


/* 泛型支持
class PeekingIterator<E> implements Iterator<E> {
    private Iterator<E> iterator;
    private E nextElement;

    public PeekingIterator(Iterator<E> iterator) {
        this.iterator = iterator;
        nextElement = iterator.next();
    }
    
    public E peek() {
        return nextElement;
    }
    
    @Override
    public E next() {
        E ret = nextElement;
        nextElement = iterator.hasNext() ? iterator.next() : null;
        return ret;
    }
    
    @Override
    public boolean hasNext() {
        return nextElement != null;
    }
}
*/

golang 解法, 执行用时: 4 ms, 内存消耗: 2.4 MB, 提交时间: 2022-11-12 12:03:07

/*   Below is the interface for Iterator, which is already defined for you.
 *
 *   type Iterator struct {
 *       
 *   }
 *
 *   func (this *Iterator) hasNext() bool {
 *		// Returns true if the iteration has more elements.
 *   }
 *
 *   func (this *Iterator) next() int {
 *		// Returns the next element in the iteration.
 *   }
 */

type PeekingIterator struct {
    iter     *Iterator
    _hasNext bool
    _next    int
}

func Constructor(iter *Iterator) *PeekingIterator {
    return &PeekingIterator{iter, iter.hasNext(), iter.next()}
}

func (it *PeekingIterator) hasNext() bool {
    return it._hasNext
}

func (it *PeekingIterator) next() int {
    ret := it._next
    it._hasNext = it.iter.hasNext()
    if it._hasNext {
        it._next = it.iter.next()
    }
    return ret
}

func (it *PeekingIterator) peek() int {
    return it._next
}

python3 解法, 执行用时: 40 ms, 内存消耗: 15 MB, 提交时间: 2022-11-12 12:02:24

# Below is the interface for Iterator, which is already defined for you.
#
# class Iterator:
#     def __init__(self, nums):
#         """
#         Initializes an iterator object to the beginning of a list.
#         :type nums: List[int]
#         """
#
#     def hasNext(self):
#         """
#         Returns true if the iteration has more elements.
#         :rtype: bool
#         """
#
#     def next(self):
#         """
#         Returns the next element in the iteration.
#         :rtype: int
#         """

class PeekingIterator:
    def __init__(self, iterator):
        """
        Initialize your data structure here.
        :type iterator: Iterator
        """
        self.iterator = iterator
        self._next = iterator.next()
        self._hasNext = iterator.hasNext()
        

    def peek(self):
        """
        Returns the next element in the iteration without advancing the iterator.
        :rtype: int
        """
        return self._next
        
        

    def next(self):
        """
        :rtype: int
        """
        ret = self._next
        self._hasNext = self.iterator.hasNext()
        self._next = self.iterator.next() if self._hasNext else 0
        return ret

    def hasNext(self):
        """
        :rtype: bool
        """
        return self._hasNext
        

# Your PeekingIterator object will be instantiated and called as such:
# iter = PeekingIterator(Iterator(nums))
# while iter.hasNext():
#     val = iter.peek()   # Get the next element but not advance the iterator.
#     iter.next()         # Should return the same value as [val].

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