LeetCode解题报告（485）-- 1630. Arithmetic Subarrays

Problem

A sequence of numbers is called arithmetic if it consists of at least two elements, and the difference between every two consecutive elements is the same. More formally, a sequence s is arithmetic if and only if s[i+1] - s[i] == s[1] - s[0] for all valid i.

For example, these are arithmetic sequences:

1
2
3

1, 3, 5, 7, 9
7, 7, 7, 7
3, -1, -5, -9

The following sequence is not arithmetic:

1	1, 1, 2, 5, 7

You are given an array of n integers, nums, and two arrays of m integers each, l and r, representing the m range queries, where the ith query is the range [l[i], r[i]]. All the arrays are 0-indexed.

Return a list of boolean elements answer, where answer[i] is true if the subarray nums[l[i]], nums[l[i]+1], ... , nums[r[i]] can be rearranged to form an arithmetic sequence, and false otherwise.

Example 1:

Input: nums = [4,6,5,9,3,7], l = [0,0,2], r = [2,3,5]
Output: [true,false,true]
Explanation:
In the 0th query, the subarray is [4,6,5]. This can be rearranged as [6,5,4], which is an arithmetic sequence.
In the 1st query, the subarray is [4,6,5,9]. This cannot be rearranged as an arithmetic sequence.
In the 2nd query, the subarray is [5,9,3,7]. This can be rearranged as [3,5,7,9], which is an arithmetic sequence.

Example 2:

1 2	Input: nums = [-12,-9,-3,-12,-6,15,20,-25,-20,-15,-10], l = [0,1,6,4,8,7], r = [4,4,9,7,9,10] Output: [false,true,false,false,true,true]

Constraints:

n == nums.length
m == l.length
m == r.length
2 <= n <= 500
1 <= m <= 500
0 <= l[i] < r[i] < n
-105 <= nums[i] <= 105

Analysis

题目给出了一个数组nums[i]，还有l和r，l[i]和r[i]组成一个区间，要求返回每一个区间是否是一个等差数列。所以这道题目的本质就是检查某个数组是否等差数列，这是一个通用的方法，在碰到其他的等差数列问题时也可以采取类似的方法。

首先找出数列的最大值mx和最小值mi，如果两者相同说明里面的元素都是一样的，等差为0，所以也是等差数列。如果两者不相等，我们尝试计算等差diff = (mx - mi) / len，但是在计算前要检查能否被整除，如果不能被整除说明不存在等差。如果存在等差的话，还不能说明这就是一个等差数列，还要逐个数字检查，检查nums[i] - mi能否整除diff，同时相同的数字还不能出现两遍（等差为0的情况之前已经判断过了，如果这里存在两个数字相同，说明不符合要求）。如果所有的检查都没有问题的话，才能认为这是一个等差数列。

Solution

无。

Code

class Solution {
public:
    vector<bool> checkArithmeticSubarrays(vector<int>& nums, vector<int>& l, vector<int>& r) {
        int size = l.size();
        vector<bool> result(size, false);
        
        for (int i = 0; i < size; ++i) {
            int left = l[i], right = r[i], len = right - left + 1;
            int mx = INT_MIN, mi = INT_MAX;
            for (int j = left; j <= right; ++j) {
                mx = max(mx, nums[j]);
                mi = min(mi, nums[j]);
            }
            
            if (mx == mi) {
                result[i] = true;
                continue;
            }
            
            if ((mx - mi) % (len - 1)) {
                continue;
            }
            vector<bool> diffs(len, false);
            int diff = (mx - mi) / (len - 1);
            int j = left;
            for (; j <= right; ++j) {
                if ((nums[j] - mi) % diff || diffs[(nums[j] - mi) / diff]) {
                    break;
                }
                diffs[(nums[j] - mi) / diff] = true;
            }
            
            if (j > right) {
                result[i] = true;
            }
        }
        return result;
    }
};

Summary

这道题目的重点就是怎么判断一个等差数列，我认为整理一下这个思路是很有用的。这道题目的分享到这里，感谢你的支持！