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?
题目标签:Design
题目链接:LeetCode / LeetCode中国
Language | Runtime | Memory |
---|---|---|
cpp | 4 ms | 1 MB |
// 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);
virtual ~Iterator();
// 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:
int pos = 0;
vector<int> vec;
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.
vec = nums;
}
// Returns the next element in the iteration without advancing the iterator.
int peek() {
return vec[pos];
}
// hasNext() and next() should behave the same as in the Iterator interface.
// Override them if needed.
int next() {
int t = peek();
pos++;
return t;
}
bool hasNext() const {
return pos < vec.size();
}
};