(Java) LeetCode 413. Arithmetic Slices —— 等差数列划分

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.

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想了半天这道题竟然是一道找规律问题…有的时候真的难以捉摸什么问题需要推导，什么问题是找规律。

如果数组是{1,2,3}，结果是1；

如果数组是{1,2,3,4}，结果是3；

如果数组是{1,2,3,4,5}，结果是6；

如果是数组是{1,2,3,4,5,6}，结果是10……

也就是说如果新扫描到的元素仍能和之前的元素保持等差数列关系，那么新形成的子等差数列个数按照1，2，3，4……的规律增加。其实找到规律这道题就做出来了。所以本题根本是寻找差子数列的终点，并以此时的结果增量更新最终结果。不满足等差数列的时候增量就要重置为1。见下文代码。

 

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Java

class Solution {
    public int numberOfArithmeticSlices(int[] A) {
        if (A == null || A.length <= 2) return 0;
        int cur = 1, res = 0, diff = A[1] - A[0];
        for (int i = 2; i < A.length; i++) {
            if (A[i] - A[i-1] == diff) res += cur++;
            else {
                diff = A[i] - A[i-1];
                cur = 1;
            }
        }
        return res;
    }   
}

 

转载于:https://www.cnblogs.com/tengdai/p/9300899.html