LeetCode Daily Question 23

May 6th 2022

5 月 6 日 每日一题

题目：284. Peeking Iterator

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 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.

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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

思路

这道题没什么好说的，看懂题就能做出来，奈何我没看懂题…

代码

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class PeekingIterator : public Iterator {
private:
    bool isHasNext;
    int nextVal;
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.
        isHasNext = Iterator::hasNext();
        if(isHasNext)
            nextVal = Iterator::next();
    }

    // Returns the next element in the iteration without advancing the iterator.
    int peek() {
        return nextVal;
    }

    // hasNext() and next() should behave the same as in the Iterator interface.
    // Override them if needed.
    int next() {
        int currNext = nextVal;
        isHasNext = Iterator::hasNext();
        if(isHasNext)
            nextVal = Iterator::next();
        return currNext;
    }

    bool hasNext() const {
        return isHasNext;
    }
};
  1. 5 月 6 日 每日一题
    1. 题目：284. Peeking Iterator
    2. 思路
    3. 代码