题干：
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.

题目解析：
1、设计struct xgr表示得到的子串，fir（第一个），last（最后一个），diff（差值）
2、先找出所有3个的子串
3、在3个的子串中慢慢扩展last直到无法扩展（根据last判断是否在t－1时是否扩展，否的话，以后都不需要考虑了）
4、count在扩展时计数

class Solution {  
public:  
   struct xgr {  
    int fir;  
    int last;  
    int diff;  

    xgr(int a, int b, int c):fir(a), last(b), diff(c){}  
};  

int numberOfArithmeticSlices(vector<int>& A) {  
    vector<xgr> re;  
    if(A.empty() || A.size() < 3)  
    return 0;  
    for (int i = 0; i < A.size() - 2; i++) {  
        if (A[i + 1] - A[i] == A[i + 2] - A[i + 1]) {  
            int fir = i;  
            int last = i + 2;  
            int diff = A[i + 1] - A[i];  

            re.push_back(xgr(fir, last, diff));  
        }  
    }  

    if (re.empty())  
        return 0;  

    int count = re.size();  
    for (int t = 4; t <= A.size(); t++) {  
        for (int j = 0; j < re.size(); j++) {  
            if ((re[j].last - re[j].fir + 1<= t - 2 )|| (re[j].last + 1 >= A.size()))  
                continue;  
            if (A[re[j].last + 1] - A[re[j].last] == re[j].diff) {  
                count++;  
                re[j].last++;  
            }  
        }  
    }  

    return count;  
}  
};