LeetCode解题报告（516）-- 900. RLE Iterator

Problem

We can use run-length encoding (i.e., RLE) to encode a sequence of integers. In a run-length encoded array of even length encoding (0-indexed), for all even i, encoding[i] tells us the number of times that the non-negative integer value encoding[i + 1] is repeated in the sequence.

• For example, the sequence arr = [8,8,8,5,5] can be encoded to be encoding = [3,8,2,5]. encoding = [3,8,0,9,2,5] and encoding = [2,8,1,8,2,5] are also valid RLE of arr.

Given a run-length encoded array, design an iterator that iterates through it.

Implement the RLEIterator class:

• RLEIterator(int[] encoded) Initializes the object with the encoded array encoded.
• int next(int n) Exhausts the next n elements and returns the last element exhausted in this way. If there is no element left to exhaust, return -1 instead.

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 rLEIterator = new RLEIterator([3, 8, 0, 9, 2, 5]); // This maps to the sequence [8,8,8,5,5].
rLEIterator.next(2); // exhausts 2 terms of the sequence, returning 8. The remaining sequence is now [8, 5, 5].
rLEIterator.next(1); // exhausts 1 term of the sequence, returning 8. The remaining sequence is now [5, 5].
rLEIterator.next(1); // exhausts 1 term of the sequence, returning 5. The remaining sequence is now [5].
rLEIterator.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.

Constraints:

• 2 <= encoding.length <= 1000
• encoding.length is even.
• 0 <= encoding[i] <= 109
• 1 <= n <= 109
• At most 1000 calls will be made to next.

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Analysis

  这是一道数组的iterator题目，题目给出一个encoding数据，长度是偶数，encoding[i]表示的是encoding[i + 1]出现的次数，所以可以理解为相邻两个数字配对出现，前一个是次数，后一个是数字本身。

  因此对于每对组合，我只需要把下标定位在第一个位置就可以得到两个信息了，而且这个下标只会指向偶数下标的位置，所以判断结束的条件是idx > size - 2。然后我们来看关键的逻辑是next(int n)这个方法，它要求的是向后遍历n个数字并返回最后的那个数字，那这里有两种情况，第一种是n比当前这个数字出现的次数要小，这个处理很容易，返回的结果就是当前的数字，然后更新下频率为arr[idx] -= n就可以了；第二种比较麻烦，因为当前这个数字的次数不足n，所以idx要往后移动，每次移动2位，移动前要把当前的频率arr[idx]置为0，同时每次移动完都要判断是否已经越界。

Solution

  无。

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Code

class RLEIterator {
public:
    RLEIterator(vector<int>& encoding) {
        arr = encoding;
        idx = 0;
    }
    
    int next(int n) {
        if (idx > arr.size() - 2) {
            return -1;
        }
        
        while (1) {
            if (n > arr[idx]) {
                n -= arr[idx];
                arr[idx] = 0;
                idx += 2;
                
                if (idx > arr.size() - 2) {
                    return -1;
                }
            } else {
                arr[idx] -= n;
                return arr[idx + 1];
            }
        }
    }
private:
    vector<int> arr;
    int idx;
};

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

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Summary

  这道题目难度比前两道要小，主要就是处理n比当前元素的频率大的情况，需要向后移动idx。这道题目的分享到这里，感谢你的支持！