900. RLE Iterator

900. RLE Iterator

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.

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没什么好讲的

class RLEIterator {

    private int pos;
    private int unused;
    private int[] arr;
    public RLEIterator(int[] A) {
        arr = A;
        pos = 0;
        unused = arr[0];
    }
    
    public int next(int n) {
        while (n > 0) {
            if (n > unused) {
                n -= unused;
                pos += 2;
                unused = 0;
                if (pos >= arr.length) return -1;
                unused = arr[pos];
            }else {
                unused -= n;
                return arr[pos + 1];
            }
        }
        
        return -1;
    }
}

/**
 * Your RLEIterator object will be instantiated and called as such:
 * RLEIterator obj = new RLEIterator(A);
 * int param_1 = obj.next(n);
 */

网上看到的一种更简洁的写法

class RLEIterator {

    private int pos;
    private int[] arr;
    public RLEIterator(int[] A) {
        arr = A;
        pos = 0;
    }
    
    public int next(int n) {
        while (pos < arr.length && n > arr[pos]) {
            n -= arr[pos];
            pos += 2;
        }
        
        if (pos >= arr.length) return -1;
        
        arr[pos] -= n;
        return arr[pos + 1];
    }
}

/**
 * Your RLEIterator object will be instantiated and called as such:
 * RLEIterator obj = new RLEIterator(A);
 * int param_1 = obj.next(n);
 */