LeetCode #413 - Arithmetic Slices - Medium

Problem

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.

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Algorithm

整理一下题意：给定一组整数，判断其中的等差数列的组数。注意等差数列最小长度为3。

首先想到的是寻找状态转移方程。对于本题，长度为n的整数序列由长度为n-1的整数数列可以得到，但是转移过程需要判断新添加的项是否与前面构成新的等差数列。若新添加的项在末尾，则需要考虑第n-1项与第n-2项、第n-3项、第n-4项…第0项是否构成新的等差数列。这个判断直观上较为复杂，因为无法得知长度为n-1的数列的最末项与前面多少项构成等差数列。

下面尝试直接枚举寻找规律

等差数列	等差数列的数目	与上一组数目的差值
123	1	1
1234	3	2
12345	6	3
123456	10	4
1234567	15	5

可以发现当前数列的数目与上一组数目的差值是成等差数列的。这一规律同样适用于相同长度不同公差的等差数列。于是在末尾每增加一项构成等差数列的数，增加的等差数列数目可通过当前数目与上一组数目的差值得到。代码如下。

//版本，时间复杂度O(n)
class Solution {
public:
    int numberOfArithmeticSlices(vector<int>& A) {
        int count = 0;
        int dd = 0;
        if(A.size()<3) return 0;

        for(int i=2;iif(A[i]-A[i-1]==A[i-1]-A[i-2]){
                dd++;
                count+=dd;
            }
            else dd = 0;
        }
        return count;
    }
};