[leetcode] 900. RLE Iterator

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 <= 1000A.length is an even integer.0 <= A[i] <= 10^9There 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