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87 lines
2.7 KiB
Markdown
87 lines
2.7 KiB
Markdown
# [284. Peeking Iterator](https://leetcode.com/problems/peeking-iterator/)
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## 题目
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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().
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**Example:**
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```
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Assume that the iterator is initialized to the beginning of the list: [1,2,3].
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Call next() gets you 1, the first element in the list.
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Now you call peek() and it returns 2, the next element. Calling next() after that still return 2.
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You call next() the final time and it returns 3, the last element.
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Calling hasNext() after that should return false.
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```
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**Follow up**: How would you extend your design to be generic and work with all types, not just integer?
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## 题目大意
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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。
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> peek() 是偷看的意思。偷偷看一看下一个元素是什么，但是并不是 next() 访问。
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## 解题思路
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- 简单题。在 PeekingIterator 内部保存 2 个变量，一个是下一个元素值，另一个是是否有下一个元素。在 next() 操作和 hasNext() 操作时，访问保存的这 2 个变量。peek() 操作也比较简单，判断是否有下一个元素，如果有，即返回该元素值。这里实现了迭代指针不移动的功能。如果没有保存下一个元素值，即没有 peek() 偷看，next() 操作继续往后移动指针，读取后一位元素。
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- 这里复用了是否有下一个元素值，来判断 hasNext() 和 peek() 操作中不移动指针的逻辑。
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## 代码
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```go
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package leetcode
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//Below is the interface for Iterator, which is already defined for you.
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type Iterator struct {
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}
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func (this *Iterator) hasNext() bool {
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// Returns true if the iteration has more elements.
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return true
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}
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func (this *Iterator) next() int {
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// Returns the next element in the iteration.
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return 0
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}
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type PeekingIterator struct {
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nextEl int
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hasEl bool
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iter *Iterator
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}
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func Constructor(iter *Iterator) *PeekingIterator {
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return &PeekingIterator{
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iter: iter,
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}
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}
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func (this *PeekingIterator) hasNext() bool {
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if this.hasEl {
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return true
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}
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return this.iter.hasNext()
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}
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func (this *PeekingIterator) next() int {
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if this.hasEl {
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this.hasEl = false
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return this.nextEl
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} else {
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return this.iter.next()
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}
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}
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func (this *PeekingIterator) peek() int {
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if this.hasEl {
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return this.nextEl
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}
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this.hasEl = true
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this.nextEl = this.iter.next()
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return this.nextEl
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}
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``` |