LeetCode: 900. RLE Iterator

题目描述

Write an iterator that iterates through a run-length encoded sequence.

The iterator is initialized by RLEIterator(int[] A), where A is a run-length encoding of some sequence. More specifically, for all even i, A[i] tells us the number of times that the non-negative integer value A[i+1] is repeated in the sequence.

The iterator supports one function: next(int n), which exhausts the next n elements (n >= 1) and returns the last element exhausted in this way. If there is no element left to exhaust, next returns -1 instead.

For example, we start with A = [3,8,0,9,2,5], which is a run-length encoding of the sequence [8,8,8,5,5]. This is because the sequence can be read as “three eights, zero nines, two fives”.

Example 1:

Input: ["RLEIterator","next","next","next","next"], [[[3,8,0,9,2,5]],[2],[1],[1],[2]]
Output: [null,8,8,5,-1]
Explanation: 
RLEIterator is initialized with RLEIterator([3,8,0,9,2,5]).
This maps to the sequence [8,8,8,5,5].
RLEIterator.next is then called 4 times:

.next(2) exhausts 2 terms of the sequence, returning 8.  The remaining sequence is now [8, 5, 5].
.next(1) exhausts 1 term of the sequence, returning 8.  The remaining sequence is now [5, 5].
.next(1) exhausts 1 term of the sequence, returning 5.  The remaining sequence is now [5].
.next(2) exhausts 2 terms, returning -1.  This is because the first term exhausted was 5,
but the second term did not exist.  Since the last term exhausted does not exist, we return -1.

Note:

0 <= A.length <= 1000
A.length is an even integer.
0 <= A[i] <= 10^9
There are at most 1000 calls to RLEIterator.next(int n) per test case.
Each call to RLEIterator.next(int n) will have 1 <= n <= 10^9.

解题思路

签到题。每次调用 next(n) 函数时，判断当前队首的元素的数量和 n 的大小关系。

• 如果队首的元素的数量大于等于 n，则将其数量减少 n 并返回该元素。
• 如果队首的元素的数量小于 n，则将 n 减少队首的元素的数量，并将该元素出队。然后继续对后面的元素做比较处理。

AC 代码

class RLEIterator {
public:
    RLEIterator(vector<int> A) : sequence(A.rbegin(), A.rend()){}
    int next(int n) {
        int lastNum = -1;

        for(int i = sequence.size()-2; i >= 0; i -= 2)
        {
            if(sequence[i+1] == 0)
            {
                sequence.pop_back();
                sequence.pop_back(); 
            }
            else if(sequence[i+1] < n && sequence[i+1] > 0)
            {
                n -= sequence[i+1];
                sequence.pop_back();
                sequence.pop_back();
            }
            else if(sequence[i+1] >= n)
            {
                lastNum = sequence[i];
                sequence[i+1] -= n;

                if(sequence[i+1] == 0)
                {
                    sequence.pop_back();
                    sequence.pop_back();
                }
                break;
            }
        }

        return lastNum;
    }

    vector<int> sequence;
};

/** * Your RLEIterator object will be instantiated and called as such: * RLEIterator obj = new RLEIterator(A); * int param_1 = obj.next(n); */