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 -1instead.

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.

这道题给了我们一种 Run-Length Encoded 的数组，就是每两个数字组成一个数字对儿，前一个数字表示后面的一个数字重复出现的次数。然后有一个 next 函数，让我们返回数组的第n个数字，题目中给的例子也很好的说明了题意。那么最暴力的方法肯定是直接还原整个数组，然后直接用坐标n去取数，但是直觉告诉我这种方法会跪，而且估计是 Memory Limit Exceeded 之类的。所以博主最先想到的是将每个数字对儿抽离出来，放到一个新的数组中。这样我们就只要遍历这个只有数字对儿的数组，当出现次数是0的时候，直接跳过当前数字对儿。若出现次数大于等于n，那么现将次数减去n，然后再返回该数字。否则用n减去次数，并将次数赋值为0，继续遍历下一个数字对儿。若循环退出了，直接返回 -1 即可，参见代码如下：

解法一：

class RLEIterator {
public:
    RLEIterator(vector<int> A) {
       for (int i = 0; i < A.size(); i += 2) {
           if (A[i] != 0) seq.push_back({A[i + 1], A[i]});
       }
    }
    
    int next(int n) {
       for (auto &p : seq) {
           if (p.second == 0) continue;
           if (p.second >= n) {
               p.second -= n;
               return p.first;
           }
           n -= p.second;
           p.second = 0;
       }
       return -1;
    }

private:
    vector<pair<int, int>> seq;
};

其实我们根本不用将数字对儿抽离出来，直接用输入数组的形式就可以，再用一个指针 cur，指向当前数字对儿的次数即可。那么在 next 函数中，我们首先来个 while 循环，判读假如 cur 没有越界，且当n大于当前当次数了，则n减去当前次数，cur 自增2，移动到下一个数字对儿的次数上。当 while 循环结束后，判断若此时 cur 已经越界了，则返回 -1，否则当前次数减去n，并且返回当前数字即可，参见代码如下：

解法二：

class RLEIterator {
public:
    RLEIterator(vector<int>& A): nums(A), cur(0) {}
    
    int next(int n) {
        while (cur < nums.size() && n > nums[cur]) {
            n -= nums[cur];
            cur += 2;
        }
        if (cur >= nums.size()) return -1;
        nums[cur] -= n;
        return nums[cur + 1];
    }
    
private:
    int cur;
    vector<int> nums;
};

Github 同步地址:

#900

参考资料：

https://leetcode.com/problems/rle-iterator/

https://leetcode.com/problems/rle-iterator/discuss/168294/Java-Straightforward-Solution-O(n)-time-O(1)-space

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