# LeetCode Arithmetic Slices

LeetCode 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.```

• n=3，抽出1个
• n=4，抽出3=1+2个，即1个长度为4的和2个长度为3的
• n=5，抽出6=1+2+3个，即1个长度为5的、2个长度为4的和3个长度为3的
• n=6，抽出10=1+2+3+4个，即...
• ...

```class Solution {
public:
inline int slice(int cnt) {
if (cnt < 3)return 0;
if (cnt == 3)return 1;
return (cnt - 1)*(cnt - 2) / 2; // 1+2+...+(cnt-2)
}
int numberOfArithmeticSlices(vector<int>& A) {
if (A.size() < 3)return 0;
int ans = 0, cnt = 2, diff = A[1] - A[0];
for (int i = 2; i < A.size(); ++i) {
if (A[i] - A[i - 1] == diff) {
++cnt;
}
else {
ans += slice(cnt);
cnt = 2;
diff = A[i] - A[i - 1];
}
}
return ans + slice(cnt);
}
};
```

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

DP一般都可以优化空间，上面的解法也可以不用一个dp数组，而只用一个变量。如下：

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