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public int numberOfArithmeticSlices(int[] A) {
    int curr = 0, sum = 0;
    for (int i=2; i<A.length; i++)
        if (A[i]-A[i-1] == A[i-1]-A[i-2]) {
            curr += 1;
            sum += curr;
        } else {
            curr = 0;
        }
    return sum;
}

def numberOfArithmeticSlices(self, A):
    """
    :type A: List[int]
    :rtype: int
    """
    opt, i = [0,0], 1
    for j in xrange(2,len(A)):
        if A[j]-A[j-1] == A[j-1]-A[j-2]:
            opt.append(opt[j-1]+i)
            i += 1
        else:
            opt.append(opt[j-1])
            i = 1
    return opt[-1]

public class Solution {
    public int numberOfArithmeticSlices(int[] A) {
        if (A == null || A.length < 3) return 0;
        int len = A.length;
        int sum = 0;
        int cur = 1;
        for (int i = 2; i < len; i++) {
            if (A[i] - A[i - 1] == A[i - 1] - A[i - 2]) {
                sum += cur;
                cur++;
            } else {
                cur = 1;
            }
        }
        return sum;
    }
}

class Solution {
public:
    int numberOfArithmeticSlices(vector<int>& A) {
        int n = A.size();
        if (n < 3) return 0;
        vector<int> dp(n, 0); // dp[i] means the number of arithmetic slices ending with A[i]
        if (A[2]-A[1] == A[1]-A[0]) dp[2] = 1; // if the first three numbers are arithmetic or not
        int result = dp[2];
        for (int i = 3; i < n; ++i) {
            // if A[i-2], A[i-1], A[i] are arithmetic, then the number of arithmetic slices ending with A[i] (dp[i])
            // equals to:
            //      the number of arithmetic slices ending with A[i-1] (dp[i-1], all these arithmetic slices appending A[i] are also arithmetic)
            //      +
            //      A[i-2], A[i-1], A[i] (a brand new arithmetic slice)
            // it is how dp[i] = dp[i-1] + 1 comes
            if (A[i]-A[i-1] == A[i-1]-A[i-2]) 
                dp[i] = dp[i-1] + 1;
            result += dp[i]; // accumulate all valid slices
        }
        return result;
    }
};