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”.
题目链接：https://leetcode.com/problems/rle-iterator/description/
Difficulty: medium
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
- 已知A存储的是每个数字及其对应出现的次数，其中0 <= A[i] <= 10^9，所以若把A还原成字符串(如[8,8,8,5,5])，这样取next是很方便，但是可能内存或者时间超限；
- 故定义index指向当前数据对，初始为0，position是遍历到当前数据对的位置，初始为0；
- 如果index超出A的长度，则返回-1；
- 如果n大于当前数据对剩余长度（当前数据对长度减去position），则index后移，n减去剩余长度，position置零；
- 反之，返回当前index对应的数字，position加n。
参考代码
class RLEIterator:
def __init__(self, A):
self.li=A
self.index=0
self.position=0
def next(self, n):
while(self.index<len(self.li)):
if(self.position + n>self.li[self.index]):
n-=(self.li[self.index]-self.position)
self.position=0
self.index+=2
else:
self.position+=n
return self.li[self.index+1]
return -1