[LeetCode] 900. RLE Iterator RLE迭代器

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.

這道題給了咱們一種 Run-Length Encoded 的數組，就是每兩個數字組成一個數字對兒，前一個數字表示後面的一個數字重複出現的次數。而後有一個 next 函數，讓咱們返回數組的第n個數字，題目中給的例子也很好的說明了題意。那麼最暴力的方法確定是直接還原整個數組，而後直接用座標n去取數，可是直覺告訴我這種方法會跪，並且估計是 Memory Limit Exceeded 之類的。因此博主最早想到的是將每一個數字對兒抽離出來，放到一個新的數組中。這樣咱們就只要遍歷這個只有數字對兒的數組，當出現次數是0的時候，直接跳過當前數字對兒。若出現次數大於等於n，那麼現將次數減去n，而後再返回該數字。不然用n減去次數，並將次數賦值爲0，繼續遍歷下一個數字對兒。若循環退出了，直接返回 -1 便可，參見代碼以下：

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;
};

其實咱們根本不用將數字對兒抽離出來，直接用輸入數組的形式就能夠，再用一個指針 cur，指向當前數字對兒的次數便可。那麼在 next 函數中，咱們首先來個 while 循環，判讀假如 cur 沒有越界，且當n大於當前當次數了，則n減去當前次數，cur 自增2，移動到下一個數字對兒的次數上。當 while 循環結束後，判斷若此時 cur 已經越界了，則返回 -1，不然當前次數減去n，而且返回當前數字便可，參見代碼以下：

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:

https://github.com/grandyang/leetcode/issues/900

參考資料：

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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