#1438. Longest Continuous Subarray With Absolute Diff Less Than or Equal to Limit

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本文介绍了一种算法，用于寻找给定整数数组中最长的非空子数组，该子数组中任意两个元素的绝对差不超过指定限制。通过示例说明了如何遍历数组并使用滑动窗口和双端队列来解决此问题，最终返回符合条件的最长子数组长度。

题目描述：

Given an array of integers nums and an integer limit, return the size of the longest non-empty subarray such that the absolute difference between any two elements of this subarray is less than or equal to limit.

Example 1:

Input: nums = [8,2,4,7], limit = 4
Output: 2 
Explanation: All subarrays are: 
[8] with maximum absolute diff |8-8| = 0 <= 4.
[8,2] with maximum absolute diff |8-2| = 6 > 4. 
[8,2,4] with maximum absolute diff |8-2| = 6 > 4.
[8,2,4,7] with maximum absolute diff |8-2| = 6 > 4.
[2] with maximum absolute diff |2-2| = 0 <= 4.
[2,4] with maximum absolute diff |2-4| = 2 <= 4.
[2,4,7] with maximum absolute diff |2-7| = 5 > 4.
[4] with maximum absolute diff |4-4| = 0 <= 4.
[4,7] with maximum absolute diff |4-7| = 3 <= 4.
[7] with maximum absolute diff |7-7| = 0 <= 4. 
Therefore, the size of the longest subarray is 2.

Example 2:

Input: nums = [10,1,2,4,7,2], limit = 5
Output: 4 
Explanation: The subarray [2,4,7,2] is the longest since the maximum absolute diff is |2-7| = 5 <= 5.

Example 3:

Input: nums = [4,2,2,2,4,4,2,2], limit = 0
Output: 3

Constraints:

1 <= nums.length <= 10^5
1 <= nums[i] <= 10^9
0 <= limit <= 10^9

class Solution {
public:
    int longestSubarray(vector<int>& nums, int limit) {
        int i = 0, j = 0;
        map<int, int> count;
        int max_length = 0;
        while (j < nums.size()) {
            count[nums[j]]++;
            while(i <= j && (count.rbegin()->first - count.begin()->first) > limit) {
                count[nums[i]]--;
                if (count[nums[i]] == 0) count.erase(nums[i]);
                i++;
            }
            max_length = max(max_length, j - i + 1);
            j++;
        }
        return max_length;
    }
};