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".

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:

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

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。

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];
    }
}