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]],,,,]
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 .

.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`.

900. RLE Iterator
``````struct RLEIterator {
prefix: Vec<usize>,
values: Vec<i32>,
index: usize,
size: usize,
}

impl RLEIterator {
fn new(a: Vec<i32>) -> Self {
let mut prefix = vec![];
let mut values = vec![];
let n = a.len();
let index = 0;
let mut prev = 0;
for i in 0..n / 2 {
let time = a[i * 2] as usize;
if time != 0 {
prev += time;
prefix.push(prev);
values.push(a[i * 2 + 1]);
}
}
let size = values.len();
RLEIterator {
prefix,
values,
index,
size,
}
}

fn next(&mut self, n: i32) -> i32 {
self.index += n as usize;
match self.prefix.binary_search(&self.index) {
Ok(i) => self.values[i],
Err(i) => {
if i < self.size {
self.values[i]
} else {
-1
}
}
}
}
}

#[test]
fn test() {
let mut obj = RLEIterator::new(vec![3, 8, 0, 9, 2, 5]);
assert_eq!(obj.next(2), 8);
assert_eq!(obj.next(1), 8);
assert_eq!(obj.next(1), 5);
assert_eq!(obj.next(2), -1);
}
``````