## Description

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:

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.

## Solutions

这里主要搞清楚题目是啥意思，Run-length Encoding 要明白是怎么回事，然后结合题意要理解，比如这里的例子：

# 输入：其实是编码后的字串
[3,8,0,9,2,5]

# 输出：编码前的字串
[8,8,8,5,5]


编码后的字串表是的含义是 [(3,8),(0,9),(2,5)]，每个括号表示的是 (数字出现次数,对应的数字)，那么题意就比较清楚了。

### 1. Stack

# Time: O(n)
# Space: O(n)
class RLEIterator:

def __init__(self, A: List[int]):
self.stack = []
n = len(A)
for i in range(0, n, 2):
if A[i] > 0:
self.stack.insert(0, (A[i], A[i+1]))

def next(self, n: int) -> int:
while n > 0:
if self.stack:
cnt, num = self.stack.pop()
if n > cnt:
n -= cnt
else:
res = num
cnt -= n
self.stack.append((cnt, num))
return res
else:
return -1

# Your RLEIterator object will be instantiated and called as such:
# obj = RLEIterator(A)
# param_1 = obj.next(n)

# 9/9 cases passed (36 ms)
# Your runtime beats 70.12 % of python3 submissions
# Your memory usage beats 100 % of python3 submissions (13.2 MB)