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 array nums.
  • int next() Returns the next element in the array and moves the pointer to the next element.
  • bool hasNext() Returns true 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 and peek are valid.
  • At most 1000 calls will be made to next, hasNext, and peek.

 

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