leetcode 900. RLE Iterator

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.

题意：每次取n个数，返回这n个数，最后的那一个。
模拟一下。

class RLEIterator {
public:
    queue<pair<int,int> > q;
    RLEIterator(vector<int> A) {
        for (int i = 0; i < A.size()-1; i+= 2) {
            int x = A[i];
            int y = A[i+1];
            //mp[y] = x;// y有x个 
            q.push({y,x});
        }
    }
    
    int next(int n) {
        while (!q.empty() && n > 0) {
            auto &x  = q.front();
            if (x.second >= n) {
                x.second -= n;
                if (x.second  == 0) q.pop();
                return x.first;
            } else {
                n -= x.second;
                q.pop();
            }
        }
        return -1;
    }
};

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