https://leetcode.com/problems/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 <= 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.
这是一个游程编码的典型题目。
该算法分为两个部分，一个是初始化，一个是调用next(n).
我们需要做的就是初始化的时候，记住这个A。 然后每次调用next(n)的时候只需要
判断n是否大于Ai
-
如果大于A[i], 那就说明不够，我们移除数组前两项，更新n，重复1
-
如果小于A[i], 则说明够了，更新A[i]
这样做，我们每次都要更新A，还有一种做法就是不更新A，而是伪更新，即用一个变量记录，当前访问到的数组位置。
很多时候我们需要原始的，那么就必须这种放了，我的解法就是这种方法。
/**
* @param {number[]} A
*/
var RLEIterator = function(A) {
this.A = A;
this.current = 0;
};
/**
* @param {number} n
* @return {number}
*/
RLEIterator.prototype.next = function(n) {
const A = this.A;
while(this.current < A.length && A[this.current] < n){
n = n - A[this.current];
this.current += 2;
}
if(this.current >= A.length){
return -1;
}
A[this.current] = A[this.current] - n; // 更新Count
return A[this.current + 1]; // 返回element
};
/**
* Your RLEIterator object will be instantiated and called as such:
* var obj = new RLEIterator(A)
* var param_1 = obj.next(n)
*/