# 900. RLE Iterator RLE 迭代器

## # 题目描述：

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 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.

## # 解题方法

``````class RLEIterator(object):

def __init__(self, A):
"""
:type A: List[int]
"""
self.A = A
self.index = 0

def next(self, n):
"""
:type n: int
:rtype: int
"""
while self.index < len(self.A) and self.A[self.index] < n:
n -= self.A[self.index]
self.index += 2
if self.index >= len(self.A):
return -1
self.A[self.index] -= n
return self.A[self.index + 1]

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

C++代码如下：

``````class RLEIterator {
public:
RLEIterator(vector<int> A) : A_(A), pos(0){
}

int next(int n) {
while (pos < A_.size() && n > 0) {
if (A_[pos] >= n) {
A_[pos] -= n;
n = 0;
} else {
n -= A_[pos];
A_[pos] = 0;
pos += 2;
}
}
if (pos >= A_.size() - 1) return -1;
return A_[pos + 1];
}
private:
vector<int> A_;
int pos;
};

/**
* Your RLEIterator object will be instantiated and called as such:
* RLEIterator obj = new RLEIterator(A);
* int param_1 = obj.next(n);
*/
``````

https://leetcode.com/problems/rle-iterator/discuss/168294/Java-Straightforward-Solution-O(n)-time-O(1)-space

## # 日期

2018 年 9 月 18 日 —— 铭记这一天 2019 年 2 月 26 日 —— 二月就要完了