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".
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.
0 <= A.length <= 1000A.lengthis an even integer.0 <= A[i] <= 10^9- There are at most
1000calls toRLEIterator.next(int n)per test case. - Each call to
RLEIterator.next(int n)will have1 <= n <= 10^9.
structRLEIterator{iterator: std::vec::IntoIter<i32>,remain:(i32,i32),}/** * `&self` means the method takes an immutable reference. * If you need a mutable reference, change it to `&mut self` instead. */implRLEIterator{fnnew(A:Vec<i32>) -> Self{Self{iterator:A.into_iter(),remain:(0, -1),}}fnnext(&mutself,mutn:i32) -> i32{while n > self.remain.0{ n -= self.remain.0;self.remain.0 = 0;matchself.iterator.next(){Some(x) => self.remain = (x,self.iterator.next().unwrap()),None => return -1,}}self.remain.0 -= n;self.remain.1}}/** * Your RLEIterator object will be instantiated and called as such: * let obj = RLEIterator::new(A); * let ret_1: i32 = obj.next(n); */