Question

413. Arithmetic Slices

中文文档

Description

An integer array 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, [1,3,5,7,9], [7,7,7,7], and [3,-1,-5,-9] are arithmetic sequences.

Given an integer array nums, return _the number of arithmetic subarrays of_ nums.

A subarray is a contiguous subsequence of the array.

 

Example 1:

Input: nums = [1,2,3,4]
Output: 3
Explanation: We have 3 arithmetic slices in nums: [1, 2, 3], [2, 3, 4] and [1,2,3,4] itself.

Example 2:

Input: nums = [1]
Output: 0

 

Constraints:

• 1 <= nums.length <= 5000
• -1000 <= nums[i] <= 1000

Solutions

Solution 1

class Solution:
  def numberOfArithmeticSlices(self, nums: List[int]) -> int:
    ans, cnt = 0, 2
    d = 3000
    for a, b in pairwise(nums):
      if b - a == d:
        cnt += 1
      else:
        d = b - a
        cnt = 2
      ans += max(0, cnt - 2)
    return ans

class Solution {
  public int numberOfArithmeticSlices(int[] nums) {
    int ans = 0, cnt = 0;
    int d = 3000;
    for (int i = 0; i < nums.length - 1; ++i) {
      if (nums[i + 1] - nums[i] == d) {
        ++cnt;
      } else {
        d = nums[i + 1] - nums[i];
        cnt = 0;
      }
      ans += cnt;
    }
    return ans;
  }
}

class Solution {
public:
  int numberOfArithmeticSlices(vector<int>& nums) {
    int ans = 0, cnt = 0;
    int d = 3000;
    for (int i = 0; i < nums.size() - 1; ++i) {
      if (nums[i + 1] - nums[i] == d) {
        ++cnt;
      } else {
        d = nums[i + 1] - nums[i];
        cnt = 0;
      }
      ans += cnt;
    }
    return ans;
  }
};

func numberOfArithmeticSlices(nums []int) (ans int) {
  cnt, d := 0, 3000
  for i, b := range nums[1:] {
    a := nums[i]
    if b-a == d {
      cnt++
    } else {
      d = b - a
      cnt = 0
    }
    ans += cnt
  }
  return
}

function numberOfArithmeticSlices(nums: number[]): number {
  let ans = 0;
  let cnt = 0;
  let d = 3000;
  for (let i = 0; i < nums.length - 1; ++i) {
    const a = nums[i];
    const b = nums[i + 1];
    if (b - a == d) {
      ++cnt;
    } else {
      d = b - a;
      cnt = 0;
    }
    ans += cnt;
  }
  return ans;
}

Solution 2

class Solution:
  def numberOfArithmeticSlices(self, nums: List[int]) -> int:
    ans = cnt = 0
    d = 3000
    for a, b in pairwise(nums):
      if b - a == d:
        cnt += 1
      else:
        d = b - a
        cnt = 0
      ans += cnt
    return ans