Title Description

If a sequence of numbers There are at least three elements , And the difference between any two adjacent elements is the same , The sequence is called the arithmetic sequence .

for example ,[1,3,5,7,9]、[7,7,7,7] and [3,-1,-5,-9] They're all arithmetic sequences .
Give you an array of integers nums , Returns an array of nums The number of subarrays of all arithmetic arrays in .

Subarray Is a continuous sequence in an array .

Their thinking

There is no idea when looking at this topic , Even read the wrong title ： The subarray is required to be a continuous sequence in the array , therefore (1,3,5) Not at all (1,2,3,4,5) Subarray . Create an array arr To hold End with each element Subarray , Because the subarray must satisfy three elements , So from the third of the array （ The index for 2） Start searching for .

Let's say the array is    (1,2,3,4,5,6,7)

• 3  --->  arr[ 2 ] = 1;   (1,2,3)
• 4  --->  arr[ 3 ] = 2;   (1,2,3,4);  (2,3,4)
• 5  --->  arr[ 4 ] = 3;  (1,2,3,4,5);  (2,3,4,5);  (3,4,5)
• 6  --->  arr[ 5 ] = 4;  (1,2,3,4,5,6);  (2,3,4,5,6);  (3,4,5,6);  (4,5,6)
• 7  --->  arr[ 6 ] = 5;  (1,2,3,4,5,6,7);  (2,3,4,5,6,7);  (3,4,5,6,7);  (4,5,6,7);  (5,6,7)

It is found that the number of subarrays found at the end of each element is exactly one more than the number of subarrays found at the end of the previous element , So the array (1,2,3,4,5,6,7) The number of subarrays of is to find the number of subarrays at the end of each element for summation . Therefore, the number of subarrays of this array is ：1+2+3+4+5=15

So the state transition equation is very clear

arr [ i ] = arr [ i -1 ] +1

And then  arr  Sum the array .

I / O example

Code

class Solution {
    public int numberOfArithmeticSlices(int[] nums) {
        int len = nums.length;
        int[] arr = new int[len];
        for(int i = 2; i < len; i++){
            if(nums[i] - nums[i - 1] == nums[i - 1] - nums[i - 2]){
                arr[i] = arr[i - 1] + 1;
            }
        }
        int sum = 0;
        for(int cnt :arr){
            sum += cnt;
        }
        return sum;
    }
}