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.
分析
题目的意思是：实现一个iterator的迭代器的next函数，我用暴力实现了一下，发现超时了。看了答案才知道代码需要动态的进行确定数的位置，如果这样的话就没什么好说的，每次来了一个n，就判断是否超出了范围，超出了就返回-1，如果没超出就要确定跳转到哪里了，如果不需要跳转就返回当前位置所对应的数就行了，如果需要跳转就要跳转后找到对应位置的数。
代码
class RLEIterator:
def __init__(self, A: List[int]):
self.A=A
self.i=0
self.q=0
def next(self, n: int) -> int:
while self.i<len(self.A):
if(self.q+n>self.A[self.i]):
n=n-(self.A[self.i]-self.q)
self.q=0
self.i+=2
else:
self.q+=n
return self.A[self.i+1]
return -1
# Your RLEIterator object will be instantiated and called as such:
# obj = RLEIterator(A)
# param_1 = obj.next(n)
参考文献
[LeetCode] Approach 1: Store Exhausted Position and Quantity
