难度: Medium
原题连接
内容描述
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:
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.
思路 1 - 时间复杂度: O(N)- 空间复杂度: O(len(A))******
周赛的时候做的这道题，刚开始很sb，抱着试试的想法直接全部装起来然后一个个返回，果然太 naive
class RLEIterator(object):
def __init__(self, A):
"""
:type A: List[int]
"""
self.lst = A
self.tmp = []
for i in range(0, len(self.lst), 2):
self.tmp.extend([self.lst[i+1]]*self.lst[i])
def next(self, n):
"""
:type n: int
:rtype: int
"""
if len(self.tmp) < n:
return -1
else:
for i in range(n):
if i == n-1:
return self.tmp.pop(0)
self.tmp.pop(0)
思路 2 - 时间复杂度: O(N)- 空间复杂度: O(len(A))******
然后写了半天，每次只取一个的逻辑，还是超时了。真的蠢。。。。
class RLEIterator(object):
def __init__(self, A):
"""
:type A: List[int]
"""
self.lst = A
self.tmp = []
self.cnt, self.num = 0, 0
def next(self, n):
"""
:type n: int
:rtype: int
"""
if self.tmp:
if n == 1:
return self.tmp.pop(0)
else:
self.tmp.pop(0)
return self.next(n-1)
else:
if self.cnt > 0:
self.tmp.append(self.num)
self.cnt -= 1
return self.next(n)
else:
if self.lst:
self.cnt, self.num = self.lst.pop(0), self.lst.pop(0)
return self.next(n)
else:
return -1
思路 3 - 时间复杂度: O(N)- 空间复杂度: O(len(A))******
利用python built-in数据结构deque的性质，直接AC, beats 100%
class RLEIterator(object):
def __init__(self, A):
"""
:type A: List[int]
"""
self._deque = collections.deque(A)
def next(self, n):
"""
:type n: int
:rtype: int
"""
while self._deque and n:
count = self._deque.popleft()
val = self._deque[0]
if count >= n:
self._deque.appendleft(count-n)
return val
else:
n -= count
self._deque.popleft()
return -1
# Your RLEIterator object will be instantiated and called as such:
# obj = RLEIterator(A)
# param_1 = obj.next(n)