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    • leetcode261 Graph Valid Tree-zh
    • leetcode262 Trips and Users-zh
    • leetcode263 ugly-number-zh
    • leetcode264 Ugly Number II-zh
    • leetcode265 Paint House II-zh
    • leetcode266 Palindrome Permutation-zh
    • leetcode267 Palindrome Permutation II-zh
    • leetcode268 Missing Number-zh
    • leetcode269 Alien Dictionary-zh
    • leetcode270 Closest Binary Search Tree Value-zh
    • leetcode271 Encode and Decode Strings-zh
    • leetcode272 Closest Binary Search Tree Value II-zh
    • leetcode273 Integer to English Words-zh
    • leetcode274 H-Index-zh
    • leetcode275 H-Index II-zh
    • leetcode276 Paint Fence-zh
    • leetcode277 Find the Celebrity-zh
    • leetcode278 First Bad Version-zh
    • leetcode279 Perfect Squares-zh
    • leetcode280 Wiggle Sort-zh
    • leetcode281 Zigzag Iterator-zh
    • leetcode282 Expression Add Operators-zh
    • leetcode283 Move Zeroes-zh
    • leetcode284 Peeking Iterator-zh
      • 题目描述
      • 解法
        • Python3
        • Java
        • ...
    • leetcode285 Inorder Successor in BST-zh
    • leetcode286 Walls and Gates-zh
    • leetcode287 Find the Duplicate Number-zh
    • leetcode288 Unique Word Abbreviation-zh
    • leetcode289 Game of Life-zh
    • leetcode290 Word Pattern-zh
  • 算法
  • leetcode-p2
geekzl.com
2021-07-19

leetcode284 Peeking Iterator-zh

# 284. 顶端迭代器 (opens new window)

English Version (opens new window)

# 题目描述

给定一个迭代器类的接口,接口包含两个方法: next() 和 hasNext()。设计并实现一个支持 peek() 操作的顶端迭代器 -- 其本质就是把原本应由 next() 方法返回的元素 peek() 出来。

示例:

假设迭代器被初始化为列表 [1,2,3]。

调用 next() 返回 1,得到列表中的第一个元素。
现在调用 peek() 返回 2,下一个元素。在此之后调用 next() 仍然返回 2。
最后一次调用 next() 返回 3,末尾元素。在此之后调用 hasNext() 应该返回 false。

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?

# 解法

# Python3

# 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.has_peeked = False
        self.peeked_element = None

    def peek(self):
        """
        Returns the next element in the iteration without advancing the iterator.
        :rtype: int
        """
        if not self.has_peeked:
            self.peeked_element = self.iterator.next()
            self.has_peeked = True
        return self.peeked_element

    def next(self):
        """
        :rtype: int
        """
        if not self.has_peeked:
            return self.iterator.next()
        result = self.peeked_element
        self.has_peeked = False
        self.peeked_element = None
        return result

    def hasNext(self):
        """
        :rtype: bool
        """
        return self.has_peeked or self.iterator.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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# Java

// 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 boolean hasPeeked;
    private Integer peekedElement;

	public PeekingIterator(Iterator<Integer> iterator) {
	    // initialize any member here.
	    this.iterator = iterator;
	}

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

	// hasNext() and next() should behave the same as in the Iterator interface.
	// Override them if needed.
	@Override
	public Integer next() {
	    if (!hasPeeked) {
            return iterator.next();
        }
        Integer result = peekedElement;
        hasPeeked = false;
        peekedElement = null;
        return result;
	}

	@Override
	public boolean hasNext() {
	    return hasPeeked || iterator.hasNext();
	}
}
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# ...


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上次更新: 2021/10/30, 12:58:38
leetcode283 Move Zeroes-zh
leetcode285 Inorder Successor in BST-zh

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