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
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,
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”.
Input: ["RLEIterator","next","next","next","next"], [[[3,8,0,9,2,5]],,,,] 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 . .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.
0 <= A.length <= 1000
A.lengthis an even integer.
0 <= A[i] <= 10^9
- There are at most
RLEIterator.next(int n)per test case.
- Each call to
RLEIterator.next(int n)will have
1 <= n <= 10^9.
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]
# 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)