题目描述:

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:

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1, 3, 5, 7, 9
7, 7, 7, 7
3, -1, -5, -9

The following sequence is not arithmetic.

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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:

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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.

这道题我首先用双指针找到所有的尽量长的连续等差数列. 对于每个数列, 假设长度为n, 那么它所包含的所有可能长度的等差数列(长度>=3)有$1+2+3+\dots+(n-2)=(n-2)(n-1)/2=(n^2-3n+2)/2$个.

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class Solution {
public:
    int numberOfArithmeticSlices(vector<int>& A) {
        if(A.size() < 3) return 0;
        int p = 0, q = 2;
        int ans = 0;
        while(q < A.size()) {
            if(A[p + 1] - A[p] != A[q] - A[p + 1]) {
                p++, q++;
                continue;
            }
            int diff = A[p + 1] - A[p];
            while(q + 1 < A.size() && A[q + 1] - A[q] == diff) q++;
            int seqLength = q - p + 1;
            ans += (seqLength * seqLength - 3 * seqLength + 2) / 2;
            p = q;
            q = p + 2;
        }
        return ans;
    }
};