(算法分析Week14)Arithmetic Slices[Medium]

413.Arithmetic Slices[Medium]

题目来源

Description

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

给出一个整数数组，求问有多少等差数列（等差数列中的数是连续的， 且一组等差数列里至少有三个数）

举个例子：

1, 2, 3 —— 1组 +1

1,2,3,4 —— 3组 +2

1,2,3,4,5 —— 6组 +3

1,2,3,4,5,6 —— 10组 +4

1,2,3,4,6 —— 3组

1,2,3,4,6,8 —— 4组 +1

1,2,3,4,6,8,10 —— 6组 +2

……

可以看到，如果新增加的数可以和原来的序列构成等差数列，那么增加的组数比上一组增加的组数多1。考虑新增的数不能构成等差数列的情况，即再次从0开始统计，按照每次+1的规律递增。

Complexity analysis

O(n)

Code

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

Result