动态规划---计算等差数列的个数

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最新推荐文章于 2024-07-29 11:00:56 发布

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分类专栏： LeetCode 文章标签：动态规划

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本文介绍了一道关于寻找数组中算术切片数量的算法题，详细解析了问题定义，并提供了C++及Python两种语言的解决方案。通过动态规划的方法高效地解决了问题。

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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、解答：先设向量dp中所有元素为0，若A[i]-A[i-1] == A[i-1][i-2] 则dp[i] = dp[i-1] + 1

3、代码

C++代码

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

Python代码

class Solution:
    def numberOfArithmeticSlices(self, A):
        """
        :type A: List[int]
        :rtype: int
        """
        dp = 0
        sum_num = 0
        for i in range(2,len(A)):
            if A[i]-A[i-1] == A[i-1]-A[i-2]:
                dp += 1
                sum_num += dp
            else:
                dp = 0
        return sum_num