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problems/900.rle-iterator.md

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@@ -69,74 +69,6 @@ Each call to RLEIterator.next(int n) will have 1 <= n <= 10^9.
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## 代码
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```js
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/*
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* @lc app=leetcode id=900 lang=javascript
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*
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* [900] RLE Iterator
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*
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* https://leetcode.com/problems/rle-iterator/description/
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*
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* algorithms
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* Medium (49.03%)
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* Total Accepted: 11.6K
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* Total Submissions: 23.5K
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* Testcase Example: '["RLEIterator","next","next","next","next"]\n[[[3,8,0,9,2,5]],[2],[1],[1],[2]]'
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*
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* Write an iterator that iterates through a run-length encoded sequence.
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*
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* The iterator is initialized by RLEIterator(int[] A), where A is a run-length
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* encoding of some sequence.  More specifically, for all even i, A[i] tells us
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* the number of times that the non-negative integer value A[i+1] is repeated
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* in the sequence.
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*
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* The iterator supports one function: next(int n), which exhausts the next n
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* elements (n >= 1) and returns the last element exhausted in this way.  If
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* there is no element left to exhaust, next returns -1 instead.
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*
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* For example, we start with A = [3,8,0,9,2,5], which is a run-length encoding
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* of the sequence [8,8,8,5,5].  This is because the sequence can be read as
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* "three eights, zero nines, two fives".
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*
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*
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*
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* Example 1:
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*
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*
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* Input: ["RLEIterator","next","next","next","next"],
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* [[[3,8,0,9,2,5]],[2],[1],[1],[2]]
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* Output: [null,8,8,5,-1]
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* Explanation:
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* RLEIterator is initialized with RLEIterator([3,8,0,9,2,5]).
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* This maps to the sequence [8,8,8,5,5].
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* RLEIterator.next is then called 4 times:
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*
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* .next(2) exhausts 2 terms of the sequence, returning 8. The remaining
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* sequence is now [8, 5, 5].
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*
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* .next(1) exhausts 1 term of the sequence, returning 8. The remaining
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* sequence is now [5, 5].
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*
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* .next(1) exhausts 1 term of the sequence, returning 5. The remaining
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* sequence is now [5].
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*
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* .next(2) exhausts 2 terms, returning -1. This is because the first term
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* exhausted was 5,
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* but the second term did not exist. Since the last term exhausted does not
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* exist, we return -1.
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*
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*
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*
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* Note:
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*
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*
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* 0 <= A.length <= 1000
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* A.length is an even integer.
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* 0 <= A[i] <= 10^9
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* There are at most 1000 calls to RLEIterator.next(int n) per test case.
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* Each call to RLEIterator.next(int n) will have 1 <= n <= 10^9.
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*
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*
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*/
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/**
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* @param {number[]} A
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*/

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