README
¶
900. RLE Iterator
题目
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.
Documentation
¶
Index ¶
Constants ¶
This section is empty.
Variables ¶
This section is empty.
Functions ¶
This section is empty.
Types ¶
type RLEIterator ¶
type RLEIterator struct {
// contains filtered or unexported fields
}
RLEIterator object will be instantiated and called as such: obj := Constructor(A); param_1 := obj.Next(n);
Source Files