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284. Peeking Iterator

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작성일 2022.04.28 19:09

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[LeetCode 시즌 3] 2022년 4월 24일 문제입니다.

https://leetcode.com/problems/peeking-iterator/

[Medium] 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<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?

#Google, #Array, #Design, #Iterator

링크

https://leetcode.com/problems/peeking-iterator/ 1015 회 연결

익명
작성일 2022.04.28 19:10

C++
Runtime: 4 ms, faster than 58.32% of C++ online submissions for Peeking Iterator.
Memory Usage: 7.5 MB, less than 72.34% of C++ online submissions for Peeking Iterator.

class PeekingIterator : public Iterator {
    vector<int> arr;
    int idx;
    
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.
	    arr = vector<int>(nums.begin(), nums.end());
        idx = -1;
	}
	
    // Returns the next element in the iteration without advancing the iterator.
	int peek() {
        return arr[idx + 1];
	}
	
	// hasNext() and next() should behave the same as in the Iterator interface.
	// Override them if needed.
	int next() {
	    return arr[++idx];
	}
	
	bool hasNext() const {
	    return idx + 1 < arr.size();
	}
};

익명
작성일 2022.04.29 22:23

문제 설명에 나와있는 대로 nums를 복사하지 않은(using only constant space) 풀이 입니다.

class PeekingIterator : public Iterator {
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.
	    if ((bNext = Iterator::hasNext())) cur = Iterator::next();
	}
	
    // Returns the next element in the iteration without advancing the iterator.
	int peek() {
        return cur;        
	}
	
	// hasNext() and next() should behave the same as in the Iterator interface.
	// Override them if needed.
	int next() {
	    auto ret = cur;
        if ((bNext = Iterator::hasNext())) cur = Iterator::next();
        return ret;
	}
	
	bool hasNext() const {
	    return bNext;
	}
    int cur;
    bool bNext;
};