算法设计与应用基础：第十二周

413. Arithmetic Slices

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Total Accepted: 19627
Total Submissions: 35813
Difficulty: Medium
Contributors:XiangyuLi926

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.

题解：题目要求要从给定的一个数列中找出子等差数列的个数，这些子数列的元素个数至少为3。先明确一点，一个形如1，2，3，4，5，6的序列有10个子等差数列，因为至少有三个元素，所以相当于是（6-2） + （6 - 2 - 1） + ... + 2 + 1 = 10。用等差数列求和公式计算即是有n个元素的等差数列有(n-2)*(n - 2 + 1) / 2个子等差数列。

于是，从头开始遍历给定的数列，当第i个元素减去第i-1个元素的差跟第i-1个元素减去i-2个元素的差相同是，计数加一。直到发现不相同的情况，则之前的子序列即为一个等差数列，用上面的公式可以计算出子等差数列的个数，然后数据重置，再从发现不同的位置重复前面的步骤，直到最后一位。把所有用公式计算出的值加起来就是最后要求的值。代码如下：