LeetCode——Arithmetic Slices

最新推荐文章于 2025-09-12 17:31:31 发布

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最新推荐文章于 2025-09-12 17:31:31 发布

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文章标签：数据结构与算法

原文链接：http://www.cnblogs.com/zhonghuasong/p/7500505.html

本文介绍了一种算法问题——寻找数组中的算术切片。通过动态规划的方法，计算相邻元素间的差值并判断连续差值是否相等，以此找出所有算术切片的数量。举例说明了算法的具体实现过程。

Question

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.

Solution

动态规划，先计算相邻之间的差值，然后遍历，每增加一个相同的差值，那么总数就加上之前相同差值的个数。

举个例子：
［1, 2, 3, 4］, 差值数组 table = [1, 1, 1]，遍历差值数组的时候，遍历到table[1]的时候，之前有一个1，那么总数加1，遍历到table[2]的时候，总数加2，因此总数求和等于1+2,因此有3组。

Code

class Solution {
public:
    int numberOfArithmeticSlices(vector<int>& A) {
        int count = 0;
        for (int i = 1; i < A.size(); i++) {
            table.push_back(A[i] - A[i - 1]);
        }
        if (table.size() < 2)
            return 0;
        
        int prev = table[0];
        int tmp = 1;
        for (int i = 1; i < table.size(); i++) {
            if (table[i] == prev) {
                count += tmp;
                tmp++;
            } else
                tmp = 1;
            prev = table[i];
        }
        
        return count;
    }
private:
    vector<int> table;    
};

转载于:https://www.cnblogs.com/zhonghuasong/p/7500505.html