413. Arithmetic Slices等差数列划分

Arithmetic Slices

题目描述

leetcode.64

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.

如果一个数列至少有三个元素，并且任意两个相邻元素之差相同，则称该数列为等差数列。

例如，以下数列为等差数列:

1, 3, 5, 7, 9
7, 7, 7, 7
3, -1, -5, -9
以下数列不是等差数列。

1, 1, 2, 5, 7

数组 A 包含 N 个数，且索引从0开始。数组 A 的一个子数组划分为数组 (P, Q)，P 与 Q 是整数且满足 0<=P<Q<N 。

如果满足以下条件，则称子数组(P, Q)为等差数组：

元素 A[P], A[p + 1], …, A[Q - 1], A[Q] 是等差的。并且 P + 1 < Q 。

函数要返回数组 A 中所有为等差数组的子数组个数。

示例:

A = [1, 2, 3, 4]

返回: 3, A 中有三个子等差数组: [1, 2, 3], [2, 3, 4] 以及自身 [1, 2, 3, 4]。

解题

如果一个数组为[1,2,3,4,5]，则三个数的等差数列有3个，四个数的等差数列有2个，五个数的等差数列有1个，因此规律是，每个大等差数列中包含的小等差数列个数规律：n个数1个，n-1个数2个…3个数n-2个。因此只要知道一共有几个大的等差数列，然后分别求每大的等差数列包含多少个小等差数列就可以了。
状态转移方程：如果是等差数列，则当前位置包含的小等差数列树为：
$dp[i]=dp[i-1]+1$

class Solution {
public:
    int numberOfArithmeticSlices(vector<int>& A) {
        int res = 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;
            }
            res += dp[i];
        }
        return res;
    }
};