900. RLE Iterator

问题：

给定一个行程长度编码序列A，即偶数为代表下一位数的个数。

如A=[3,8,0,9,2,5]，是序列【88855】的编码。

next(n)函数返回，对被编码序列消化n个数后最后消化的数值。实现初始化函数RLEIterator和next函数。

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:
0 <= A.length <= 1000
A.length is an even integer.
0 <= A[i] <= 10^9
There are at most 1000 calls to RLEIterator.next(int n) per test case.
Each call to RLEIterator.next(int n) will have 1 <= n <= 10^9.

　　

解法：

该问题是顺序消化序列，则使用queue来实现。

而queue对于元素修改是不利的，因此再使用vector来存储数据作为字典，queue来存取index即可。

 

代码参考：

class RLEIterator {
public:
    queue<int> Seq;
    vector<int> AA;
    RLEIterator(vector<int>& A) {
        AA=A;
        for(int i=1; i<A.size(); i+=2){
            if(A[i-1]!=0){
                Seq.push(i);
            }
        }
    }
    
    int next(int n) {
        while(!Seq.empty()){
            int p=Seq.front();
            if(AA[p-1]>=n){
                AA[p-1]-=n;
                if(AA[p-1]==0) Seq.pop();
                return AA[p];
            }else{
                n-=AA[p-1];
                Seq.pop();
            }
        }
        return -1;
    }
};

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