# 900. RLE-Iterator

difficulty: Medium

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".

&#x20;

**Example 1:**

<pre><code><strong>Input: ["RLEIterator","next","next","next","next"], [[[3,8,0,9,2,5]],[2],[1],[1],[2]]
</strong><strong>Output: [null,8,8,5,-1]
</strong><strong>Explanation: 
</strong>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.
</code></pre>

**Note:**

1. `0 <= A.length <= 1000`
2. `A.length` is an even integer.
3. `0 <= A[i] <= 10^9`
4. There are at most `1000` calls to `RLEIterator.next(int n)` per test case.
5. Each call to `RLEIterator.next(int n)` will have `1 <= n <= 10^9`.

## Method One

```Java
class RLEIterator {
    int[] A;
    int index;
    int count;
    public RLEIterator(int[] A) {
        this.A = A;
        this.index = 0;
        this.count = 0;
        
    }
    
    public int next(int n) {
        while(index < A.length){
            if( count + n > A[index]) {
                n -= A[index] - count;
                index += 2;
                count = 0;
            }else{
                count += n;
                return A[index + 1];
            }
        }
        return -1;
    }
}
​
/**
 * Your RLEIterator object will be instantiated and called as such:
 * RLEIterator obj = new RLEIterator(A);
 * int param_1 = obj.next(n);
 */
​
```

2023 写的，上面的不是很确定是不是抄的答案。

```java
class RLEIterator {
    private int[] nums;
    private int cur;
    public RLEIterator(int[] encoding) {
        this.nums = encoding;
        this.cur = 0;
    }
    
    public int next(int n) {
        while (cur < nums.length - 1 && n > nums[cur]) {
            n -= nums[cur];
            nums[cur] = 0;
            cur += 2;
        }
        if (cur >= nums.length) return -1;
        nums[cur] -= n;
        return nums[cur + 1];
    }
}
```
