Problem Peeking Iterator
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().
Here is an example. Assume that the iterator is initialized to the beginning of the list: [1, 2, 3]
.
Call next()
gets you 1, the first element in the list.
Now you call peek()
and it returns 2, the next element. Calling next()
after that still return 2.
You call next()
the final time and it returns 3, the last element. Calling hasNext()
after that should return false.
Hint:
- Think of "looking ahead". You want to cache the next element.
- Is one variable sufficient? Why or why not?
- Test your design with call order of
peek()
beforenext()
vsnext()
beforepeek()
. - For a clean implementation, check out Google's guava library source code.
Follow up: How would you extend your design to be generic and work with all types, not just integer?
Credits:
Special thanks to @porker2008 for adding this problem and creating all test cases.
peek() 需要用Iterator::next() 来实现,可是又不想move to the next element, 所以用一个数来记录当前的peek element , 同时用movedToNext来记录这个数是否有效。
调用peek()时,如果movedToNext是有效的,那再调用peek()时只需返回 val. 如果是无效的,那就调用next() 来使其有效, val 来记录next()返回值。
// 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 {
bool movedToNext;
int val;
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.
movedToNext = false;
}
// Returns the next element in the iteration without advancing the iterator.
int peek() {
if(movedToNext) return val;
movedToNext = true;
val = Iterator::next();
return val;
}
// hasNext() and next() should behave the same as in the Iterator interface.
// Override them if needed.
int next() { // easy to be buggy
if(movedToNext) {
movedToNext = false;
return val;}
return Iterator::next();
}
bool hasNext() const {
if(movedToNext) return true;
return Iterator::hasNext();
}
};