LeetCode 284 Peeking Iterator

LeetCode

Design an iterator that supports the peek operation on an existing iterator in addition to the hasNext and the next operations.

Implement the PeekingIterator class:

PeekingIterator(Iterator<int> nums) Initializes the object with the given integer iterator iterator. int next() Returns the next element in the array and moves the pointer to the next element.
boolean hasNext() Returns true if there are still elements in the array. int peek() Returns the next element in the array without moving the pointer.

Note: Each language may have a different implementation of the constructor and Iterator, but they all support the int next() and boolean hasNext() functions.

Example 1:

Input
[“PeekingIterator”, “next”, “peek”, “next”, “next”, “hasNext”]
[[[1, 2, 3]], [], [], [], [], []]
Output
[null, 1, 2, 2, 3, false]

Explanation
PeekingIterator 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?

题目大意：

实现数组的peeking Iterator。数组的Iterator是给定的。

解题思路：

预读一个数

解题步骤：

N/A

注意事项：

1. 预读一个next数

Python代码：

class PeekingIterator(TestCases):

    def __init__(self, iterator):
        """
        Initialize your data structure here.
        :type iterator: Iterator
        """
        self.iterator = iterator
        if self.iterator.hasNext():
            self.num = self.iterator.next()
        else:
            self.num = None

    def peek(self):
        """
        Returns the next element in the iteration without advancing the iterator.
        :rtype: int
        """
        return self.num

    def next(self):
        """
        :rtype: int
        """
        tmp = self.num
        if self.iterator.hasNext():
            self.num = self.iterator.next()
        else:
            self.num = None
        return tmp

    def hasNext(self):
        """
        :rtype: bool
        """
        return self.num is not None

算法分析：

时间复杂度为O(1)，空间复杂度O(1)