至少有三个数，任意两个之间的差值相等 Arithmetic Slices

问题：

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.

解决：

① 动态规划，dp[i]表示当前知足arithmetic slices的三个连续的值的个数，而总的个数是知足条件的个数的和。

class Solution {//2ms
    public int numberOfArithmeticSlices(int[] A) {
        if (A.length < 3) return 0;
        int p = 2;
        int len = A.length;
        int[] dp = new int[len];
        int res = 0;
        while(p < len){
            if (A[p] - A[p - 1] == A[p - 1] - A[p - 2]){//只要差值相等便可，而不是差值的绝对值
                dp[p] = dp[p - 1] + 1;
            }
            res += dp[p];
            p ++;
        }
        return res;
    }
}

② 只使用一个一个变量记录局部个数。

class Solution { //2ms     public int numberOfArithmeticSlices(int[] A) {         if (A.length < 3) return 0;         int p = 2;         int len = A.length;         int res = 0;         int cur = 0;         while(p < len){             if (A[p] - A[p - 1] == A[p - 1] - A[p - 2]){                 cur += 1;                 res += cur;             }else {                 cur = 0;             }             p ++;         }         return res;     } }