Leetcode[413] Arithmetic Slices

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最新推荐文章于 2025-08-15 16:44:13 发布

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文章标签：数据结构与算法

原文链接：https://segmentfault.com/a/1190000007510145

本文介绍了解决LeetCode第413题“Arithmetic Slices”的方法，该题旨在找出数组中所有等差序列的个数。通过动态规划的方式，当数组中的三个连续元素构成等差数列时进行计数，并累加得到最终答案。

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Leetcode[413] Arithmetic Slices

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.

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

DP

复杂度
O(N)

思路
找数组里面的等差数列的个数。想法是如果一开始三个数就满足是等差数列的话，就在当前已有的数目上不断的累加count的结果。然后更新sum。

代码

public int numberOfArithmeticSlices(int[] nums) {
    int cur = 0, sum = 0;
    for(int i = 2; i < nums.length; i ++) {
        if(nums[i] == nums[i - 1] && nums[i - 1] == nums[i - 1]) {
            cur += 1;
            sum += cur;
        }
        else {
            cur = 0;
        }
    }
    return sum;
}