# 0413. Arithmetic Slices
###### tags: `Leetcode` `Medium` `Bloomberg` `Arithmetic Sequence`
Link: https://leetcode.com/problems/arithmetic-slices/
## 思路 $O(N)$ $O(1)$
sliding window
每次先找到最长的subsequence，如果长度大于等于3，那么就开始计算它包含几个有效的subsequence
计算方法：
假设subsequence长度为a（a>=3），则长度为a的答案有1个，a-1的有2个，...， 3的有（a+1）-3个，所以总共有（a-3+1)(a-3+1+1)个
## Code

class Solution {
    public int numberOfArithmeticSlices(int[] nums) {
        int left = 0;
        int right = 1;
        int ans = 0;
        while(right < nums.length){
            int diff = nums[right]-nums[left];
            while(right+1<nums.length && nums[right]+diff == nums[right+1]){
                right++;
            }
            if(right-left+1>=3){
                ans += (right-left+1-3+1)*(right-left+1-3+1+1)/2;
            }
            left = right;
            right++;
        }
        return ans;
    }
}