動機
要多翻一格
Problem
Design an iterator that supports the peek
operation on a list in addition to the hasNext
and the next
operations.
Implement the PeekingIterator
class:
PeekingIterator(int[] nums)
Initializes the object with the given integer arraynums
.int next()
Returns the next element in the array and moves the pointer to the next element.bool hasNext()
Returnstrue
if there are still elements in the array.int peek()
Returns the next element in the array without moving the pointer.
Example 1:
Input[PeekingIterator, next, peek, next, next, hasNext][[[1, 2, 3]], [], [], [], [], []]Output[null, 1, 2, 2, 3, false]ExplanationPeekingIterator peekingIterator = new PeekingIterator([1, 2, 3]); // [1,2,3]peekingIterator.next(); // return 1, the pointer moves to the next element [1,2,3].peekingIterator.peek(); // return 2, the pointer does not move [1,2,3].peekingIterator.next(); // return 2, the pointer moves to the next element [1,2,3]peekingIterator.next(); // return 3, the pointer moves to the next element [1,2,3]peekingIterator.hasNext(); // return False
Constraints:
1 <= nums.length <= 1000
1 <= nums[i] <= 1000
- All the calls to
next
andpeek
are valid. - At most
1000
calls will be made tonext
,hasNext
, andpeek
.
Follow up: How would you extend your design to be generic and work with all types, not just integer?
Sol
class PeekingIterator:
def __init__(self, iterator):
self.l = iterator
self.prev,self.now = None, self.l.next()
def peek(self):
return self.now
def next(self):
if self.l.hasNext():
self.prev, self.now = self.now, self.l.next()
else:
self.prev, self.now = self.now, None
return self.prev
def hasNext(self):
return self.now is not None