## 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:**

`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`

.

## 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)
```