1 题目

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.

2 尝试解

2.1 分析

在整形向量A中，求由若干连续元素组成的所有的等差数列的个数。

只需要找出A中所有的尽可能连续的等差数列，然后对于每一个等差数列，求一个组合即可。

2.2 代码

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

3 标准解

public class Solution {
    public int NumberOfArithmeticSlices(int[] nums) {
        var n = nums.Length;

        var result = 0;

        var dp = new int[n];
        for (int i = 2; i < n; i++) {
            if (nums[i - 1] - nums[i - 2] == nums[i] - nums[i - 1]) {
                dp[i] = dp[i - 1] + 1;
            }
            result += dp[i];
        }
        return result;
    }
}