題目：

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.

解釋：
動態規劃
解法1：
用DP來做，定義一個一維dp陣列，其中dp[i]表示，到i位置為止的Arithmetic Slices的個數，那麼我們從第三個數字開始遍歷，如果當前數字和之前兩個數字構成Arithmetic Slices，那麼我們更新dp[i]為dp[i-1]+1
dp[i]=dp[i-1]+1可以理解為，當A[i]可以和A[i-1]以及A[i-2]

class Solution(object):
    def numberOfArithmeticSlices(self, A):
        """
        :type A: List[int]
        :rtype: int
        """
        #解法1：
        len_A=len(A)
        if len_A<3:
            return 0
        dp=[0]*len_A
        result=0
        for i in xrange(2,len_A):
            if 2*A[i-1]==A[i-2]+A[i]:
                dp[i]=dp[i-1]+1
            result+=dp[i]
        return result 

class Solution(object):
    def numberOfArithmeticSlices(self, A):
        """
        :type A: List[int]
        :rtype: int
        """
        #解法2：
        len_A=len(A)
        if len_A<3:
            return 0
        cur=0
        result=0
        for i in xrange(2,len_A):
            if A[i-1]*2==A[i]+A[i-2]:
                cur+=1
                result+=cur
            else:
                cur=0
        return result

c++程式碼：

class Solution {
public:
    int numberOfArithmeticSlices(vector<int>& A) {
        //解法1
        int len_A=A.size();
        if (len_A<3)
            return 0;
        int result=0;
        vector<int>dp(len_A,0);
        for (int i =2;i<len_A;i++)
        {
            if(2*A[i-1]==A[i-2]+A[i])
                dp[i]=dp[i-1]+1;
            result+=dp[i];
        }
        return result;
    }
};

class Solution {
public:
    int numberOfArithmeticSlices(vector<int>& A) {
        //解法2
        int len_A=A.size();
        if (len_A<3)
            return 0;
        int result=0;
        int cur=0;
        for (int i =2;i<len_A;i++)
        {
            if(2*A[i-1]==A[i-2]+A[i])
                cur+=1;
            else 
                cur=0;
            result+=cur;
        }
        return result;
    }
};

總結：