【python3】leetcode 900. RLE Iterator (Medium)
阿新 • • 發佈:2019-01-06
Write an iterator that iterates through a run-length encoded sequence.
The iterator is initialized by
RLEIterator(int[] A)
, whereA
is a run-length encoding of some sequence. More specifically, for all eveni
,A[i]
tells us the number of times that the non-negative integer valueA[i+1]
is repeated in the sequence.The iterator supports one function:
next(int n)
, which exhausts the nextn
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.
class RLEIterator(object):
def __init__(self, A):
"""
:type A: List[int]
"""
self.A = A
self.sum = 0
for i in range(0,len(A),2):
self.sum += A[i]
def next(self, n):
"""
:type n: int
:rtype: int
"""
if n > self.sum:
self.sum = 0
return -1
else:
while(n > self.A[0]):
n -= self.A[0]
self.sum -= self.A[0]
del self.A[0:2]
self.sum -= n
self.A[0] -= n
return self.A[1]
# Your RLEIterator object will be instantiated and called as such:
# obj = RLEIterator(A)
# param_1 = obj.next(n)