LeetCode: Arithmetic Slices

最新推荐文章于 2020-04-17 15:01:37 发布

原创最新推荐文章于 2020-04-17 15:01:37 发布 · 247 阅读

CC 4.0 BY-SA版权

本文介绍了一种高效算法，用于计算给定数组中等差数列切片的数量。通过一次遍历即可完成计算，时间复杂度为O(n)。

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.

题目定义的Arithmetic，其实就是长度至少为3的等差数列。题目的要求即为找出一个数组中Arithmetic的个数。
发现一个规律：总长度为3的Arithmetic只有一个Arithmetic Slices，然后Arithmetic的总长度每增加一（第i次增加，增加后Arithmetic的长度为3+i），Arithmetic Slices的个数就增加i+1。所以，对于长度为n（n>=3）的Arithmetic，Arithmetic Slices的个数就为:
1+（4-3+1）+(5-3+1)+…+(n-3+1）=1+2+3+…+(n-2)=(n-1)(n-2)/2（n>=3）

所以，从前往后遍历数组（因为一个Arithmetic的长度至少为3，可以从第3个元素开始遍历，因为前面不可能存在Arithmetic），找出若干个最长的Arithmetic，从而计算Arithmetic Slice的个数。算法的时间复杂度为o(n)，Accepted的代码：

class Solution {
public:
    int numberOfArithmeticSlices(vector<int>& A) {
        int count=0;
        int size=2;

        for(int i=2;i<A.size();i++)
        {
            if(A[i]-A[i-1]==A[i-1]-A[i-2]) size++;
            else
            {
                count+=(size-1)*(size-2)/2;
                size=2;
            }
        }
        if(size!=2) count+=(size-1)*(size-2)/2;//最后一个元素在某个Arithmetic中
        return count;
    }
};