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 iA[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:

  1. 0 <= A.length <= 1000
  2. A.length is an even integer.
  3. 0 <= A[i] <= 10^9
  4. There are at most 1000 calls to RLEIterator.next(int n) per test case.
  5. Each call to RLEIterator.next(int n) will have 1 <= n <= 10^9.

Runtime: 95 ms, faster than 82.10% of Java online submissions for RLE Iterator.

class RLEIterator {
private Queue<Integer> q = new ArrayDeque<>();
private int cnt = -;
private int value = -;
public RLEIterator(int[] A) {
for(int i=; i<A.length; i++){
q.add(A[i]);
}
} public int next(int n) {
if(cnt == -){
if(q.isEmpty()) {
return -;
}else {
cnt = q.peek();q.poll();
value = q.peek(); q.poll();
}
}
if(cnt >= n){
cnt -= n;
return value;
}else {
n -= cnt;
cnt = -;
}
while(!q.isEmpty()){
cnt = q.peek(); q.poll();
value = q.peek(); q.poll();
if(cnt >= n) {
cnt -= n;
return value;
}else{
n -= cnt;
cnt = -;
}
}
return -;
}
}

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