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.

代码一

class Solution {

public:

    int numberOfArithmeticSlices(vector<int>& A) {

        if(A.size()<=2) return 0;

        int cnt=0;

        for(int i=0;i<=A.size()-3;i++){

            int temp=A[i+1]-A[i];

            for(int j=i+2;j<A.size();j++)

                if(A[j]-A[j-1]==temp)

                    cnt++;

                else

                    break;

        }

        return cnt;

    }

};

代码二：动态规划

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;

    }

};