LC 413. Arithmetic Slices

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.

Runtime: 4 ms, faster than 35.11% of C++ online submissions for Arithmetic Slices.

这一题的slice是连续的，所以只要判断i, i-1, i-2的关系然后递推就行了。

dp[i] = dp[i-1] + 1, 这里的dp[i]是以index i结尾的等差数列。 意思是，如果当前的位置能和之前两个位置组成等差数列，那么该位置上之前的的等差数列就等于

之前一个位置上结尾的(dp[i-1])数量（因为如果之前的位置上有相同差值的等差数列，再连接上以index i结尾的等差数列，还是可行的）然后再加上1，这个1

是以当前位置作为结尾，用前面的两个数合并起来的一个新的三个数的等差数列，这在之前的情况中是没有出现的。

class Solution {
public:
    int numberOfArithmeticSlices(vector<int>& A) {
      int ret = , cur = , n = A.size();
      for(int i=; i<n; i++){
        if(A[i] - A[i-] == A[i-] - A[i-]){
          cur += ;
          ret += cur;
        }else {
          cur = ;
        }
      }
      return ret;
    }
};