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.

思路：就是要注意，前面是表示fre，後面是value；簡潔的版本的做法是直接modify A數組的頻率，如果相等的時候，直接歸零，return value，n > fre，跳過；

class RLEIterator {

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

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

 我自己寫的版本是記錄fre, value in node class, 然後用queue去pop；

class RLEIterator {
    private class Node {
        int value;
        int fre;
        public Node(int value, int fre) {
            this.value = value;
            this.fre = fre;
        }
    }

    Queue<Node> queue;
    public RLEIterator(int[] A) {
        queue = new LinkedList<Node>();
        convert(queue, A);
    }
    
    private void convert(Queue<Node> queue, int[] A) {
        for(int i = 0; i < A.length - 1; i += 2) {
            int fre = A[i];
            int value = A[i + 1];
            if(fre > 0) {
                queue.offer(new Node(value, fre));
            }
        }
    }
    
    public int next(int n) {
        int res = -1;
        while(n > 0 && !queue.isEmpty()) {
            Node node = queue.peek();
            if(n == node.fre) {
                queue.poll();
                n = 0;
                res = node.value;
            } else if(n > node.fre) {
                n -= node.fre;
                queue.poll();
            } else {
                // n < node.fre;
                node.fre -= n;
                n = 0;
                res = node.value;
            }
        }
        if(n > 0 && queue.isEmpty()) {
            return -1;
        }
        return res;
    }
}

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