title

problem link

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

solution

We can store an index i and quantity q which represents that q elements of A[i] (repeated A[i+1] times) are exhausted.

For example, if we have A = [1,2,3,4] (mapping to the sequence [2,4,4,4]) then i = 0, q = 0 represents that nothing is exhausted; i = 0, q = 1 represents that [2] is exhausted, i = 2, q = 1 will represent that we have currently exhausted [2, 4], and so on.

q记录了上次n的位置

class RLEIterator {
public:
	int i, q;
	vector<int> A;
	RLEIterator(vector<int> A) {
		this->A = A;
		i = 0;
		q = 0;
	}

	int next(int n) {
		while (i < A.size()) {
			//n+q超过当前值， q重新赋值向前移动， i每次都是偶数，也就是个数3x8的3
			if (n + q > A[i]) {
				n -= A[i]-q;
				q = 0;
				i += 2;
			}
			else {
				q += n;
				return A[i + 1];
			}

		}
		return -1;
	}
};

class RLEIterator:
    def __init__(self, A):
        """
        :type A: List[int]
        """
        self.A = A
        self.p = 0

    def next(self, n):
        """
        :type n: int
        :rtype: int
        """
        while self.p < len(self.A) and n > self.A[self.p]:
            n -= self.A[self.p]
            self.p += 2
        
        if self.p >= len(self.A):
            return -1
        
        self.A[self.p] -= n
        return self.A[self.p+1]