算法分析与设计课程13——413. Arithmetic Slices-优快云博客

本文链接：https://blog.youkuaiyun.com/Brother_Xiang/article/details/72860920

本文介绍了一种用于计算数组中算术切片数量的方法。通过遍历数组并检查每三个连续元素之间的等差特性来确定算术切片的存在。算法能够有效地计算出所有可能的算术序列。

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一、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.

二、分析

遍历一遍数组就能得到答案，在遍历数组，取三个连续数字a1,a2,a3判断是否符合条件，如果符合条件，再判断a2,a3,a4是否符合条件，如果符合条件，则a2,a3,a4计数为2（即其本身和a1,a2,a3,a4），同理，a3,a4,a5计数为3,则a1,a2,a3,a4,a5计数为1+2+3=6，据此完成以下代码

三、源代码

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