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.

方法1：Dynamic Programming

这个题目一看是求方案个数，就想到了动归

  1. state: dp[i] means the number of arithmetic slices ends with A[i]
  2. function: dp[i] = dp[i - 1] + 1. A[i] 跟dp[i - 1]中的每一个方案都成再组成一个数列，但是增加了一个包含A[i]的长度为3的数列。
  3. init:
  4. result: sum of dp[i]
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public int numberOfArithmeticSlices(int[] A) {
    if (A == null || A.length < 3)
        return 0;
    int[] dp = new int[A.length];
    int ans = 0;
    for (int i = 2; i < A.length; i++) {
        if (A[i] - A[i - 1] == A[i - 1] - A[i - 2]) {
            dp[i] = dp[i - 1] + 1;
        }
        ans += dp[i];
    }
    return ans;
}

方法2：Math

这里用到了一个数学公式，长度为n的数列，长度>=3的连续子数列有(n - 1) * (n - 2) / 2个。

一个长度为n的数列n (n >= 3), 其中有长度为3的子数列n - 2个, 长度为4的子数列n - 3个，以此类推，直到1. 这是一个等差数列。

按照等差数列求和公式：sum = (a1 + an) n / 2. 依本题代入 (1 + n - 2) (n - 2) / 2 = (n - 1) * (n - 2) / 2.

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public int numberOfArithmeticSlices(int[] A) {
    if (A == null || A.length < 3)
        return 0;
    int sum = 0;
    int len = 2;
    for (int i = 2; i < A.length; i++) {
        if (A[i] - A[i - 1] == A[i - 1] - A[i - 2]) {
            len++;
        } else {
            if (len > 2) {
                sum += calculateSlices(len);
            }
            // reset the length of new slice
            len = 2;
        }
    }
    // add up the slice in the rear
    if (len > 2)
        sum += calculateSlices(len);
    return sum;
}
private int calculateSlices(int n) {
    return (n - 1) * (n - 2) / 2;
}