至少有三個數，任意兩個之間的差值相等 Arithmetic Slices

問題：

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.

解決：

① 動態規劃，dp[i]表示當前知足arithmetic slices的三個連續的值的個數，而總的個數是知足條件的個數的和。

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，而不是差值的絕對值
                dp[p] = dp[p - 1] + 1;
            }
            res += dp[p];
            p ++;
        }
        return res;
    }
}

② 只使用一個一個變量記錄局部個數。

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;     } }