LeetCode【413】Arithmetic Slices

最新推荐文章于 2020-07-17 16:40:39 发布

原创最新推荐文章于 2020-07-17 16:40:39 发布 · 376 阅读

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本文介绍了一种高效计算数组中所有算术切片数量的方法。通过定义算术切片并给出示例，阐述了如何利用数学知识简化动态规划过程，实现对算术序列的有效识别。

A sequence of number is called arithmetic if it consists of at least three elements and if the difference between any two consecutive elements is the same.

For example, these are arithmetic sequence:

1, 3, 5, 7, 9
7, 7, 7, 7
3, -1, -5, -9
The following sequence is not arithmetic.

1, 1, 2, 5, 7

A zero-indexed array A consisting of N numbers is given. A slice of that array is any pair of integers (P, Q) such that 0 <= P < Q < N.

A slice (P, Q) of array A is called arithmetic if the sequence:
A[P], A[p + 1], …, A[Q - 1], A[Q] is arithmetic. In particular, this means that P + 1 < Q.

The function should return the number of arithmetic slices in the array A.

Example:

A = [1, 2, 3, 4]

return: 3, for 3 arithmetic slices in A: [1, 2, 3], [2, 3, 4] and [1, 2, 3, 4] itself.

这道题是被归到了Dynamic Programming 里，在解题中也确实可以用，不过我稍微用了一些数学上关于数列的知识，便可以更加易于理解。

class Solution {
public:
    int numberOfArithmeticSlices(vector<int>& A) {
        int begin = 0, end =0;
        int count = 0;
        for(;begin<A.size()&&end<A.size();) {
            begin = end;
            int t = findLong(begin, end, A);
            count += (t*(t+1))/2;

        }
        return count;
    }
    int findLong(int begin, int & end, vector<int>& A){
        int count = 0;
        while(1)
        {
            if(begin==A.size()-2) 
            {
                end = begin+1;
                return count;
            }
            if(A[begin]+A[begin+2]==2*A[begin+1])count++;
            else break;
            begin++;
        }
        end = begin+1;
        return count;
    }
};